multithreading – How to get the return value from a thread in python?

The Question :

389 people think this question is useful

The function foo below returns a string 'foo'. How can I get the value 'foo' which is returned from the thread’s target?

from threading import Thread

def foo(bar):
    print('hello {}'.format(bar))
    return 'foo'

thread = Thread(target=foo, args=('world!',))
thread.start()
return_value = thread.join()

The “one obvious way to do it”, shown above, doesn’t work: thread.join() returned None.

The Question Comments :

The Answer 1

98 people think this answer is useful

In Python 3.2+, stdlib concurrent.futures module provides a higher level API to threading, including passing return values or exceptions from a worker thread back to the main thread:

import concurrent.futures

def foo(bar):
    print('hello {}'.format(bar))
    return 'foo'

with concurrent.futures.ThreadPoolExecutor() as executor:
    future = executor.submit(foo, 'world!')
    return_value = future.result()
    print(return_value)

The Answer 2

287 people think this answer is useful

FWIW, the multiprocessing module has a nice interface for this using the Pool class. And if you want to stick with threads rather than processes, you can just use the multiprocessing.pool.ThreadPool class as a drop-in replacement.

def foo(bar, baz):
  print 'hello {0}'.format(bar)
  return 'foo' + baz

from multiprocessing.pool import ThreadPool
pool = ThreadPool(processes=1)

async_result = pool.apply_async(foo, ('world', 'foo')) # tuple of args for foo

# do some other stuff in the main process

return_val = async_result.get()  # get the return value from your function.

The Answer 3

277 people think this answer is useful

One way I’ve seen is to pass a mutable object, such as a list or a dictionary, to the thread’s constructor, along with a an index or other identifier of some sort. The thread can then store its results in its dedicated slot in that object. For example:

def foo(bar, result, index):
    print 'hello {0}'.format(bar)
    result[index] = "foo"

from threading import Thread

threads = [None] * 10
results = [None] * 10

for i in range(len(threads)):
    threads[i] = Thread(target=foo, args=('world!', results, i))
    threads[i].start()

# do some other stuff

for i in range(len(threads)):
    threads[i].join()

print " ".join(results)  # what sound does a metasyntactic locomotive make?

If you really want join() to return the return value of the called function, you can do this with a Thread subclass like the following:

from threading import Thread

def foo(bar):
    print 'hello {0}'.format(bar)
    return "foo"

class ThreadWithReturnValue(Thread):
    def __init__(self, group=None, target=None, name=None,
                 args=(), kwargs={}, Verbose=None):
        Thread.__init__(self, group, target, name, args, kwargs, Verbose)
        self._return = None
    def run(self):
        if self._Thread__target is not None:
            self._return = self._Thread__target(*self._Thread__args,
                                                **self._Thread__kwargs)
    def join(self):
        Thread.join(self)
        return self._return

twrv = ThreadWithReturnValue(target=foo, args=('world!',))

twrv.start()
print twrv.join()   # prints foo

That gets a little hairy because of some name mangling, and it accesses “private” data structures that are specific to Thread implementation… but it works.

For python3

class ThreadWithReturnValue(Thread):
    def __init__(self, group=None, target=None, name=None,
                 args=(), kwargs={}, Verbose=None):
        Thread.__init__(self, group, target, name, args, kwargs)
        self._return = None
    def run(self):
        print(type(self._target))
        if self._target is not None:
            self._return = self._target(*self._args,
                                                **self._kwargs)
    def join(self, *args):
        Thread.join(self, *args)
        return self._return

The Answer 4

88 people think this answer is useful

Jake’s answer is good, but if you don’t want to use a threadpool (you don’t know how many threads you’ll need, but create them as needed) then a good way to transmit information between threads is the built-in Queue.Queue class, as it offers thread safety.

I created the following decorator to make it act in a similar fashion to the threadpool:

def threaded(f, daemon=False):
    import Queue

    def wrapped_f(q, *args, **kwargs):
        '''this function calls the decorated function and puts the 
        result in a queue'''
        ret = f(*args, **kwargs)
        q.put(ret)

    def wrap(*args, **kwargs):
        '''this is the function returned from the decorator. It fires off
        wrapped_f in a new thread and returns the thread object with
        the result queue attached'''

        q = Queue.Queue()

        t = threading.Thread(target=wrapped_f, args=(q,)+args, kwargs=kwargs)
        t.daemon = daemon
        t.start()
        t.result_queue = q        
        return t

    return wrap

Then you just use it as:

@threaded
def long_task(x):
    import time
    x = x + 5
    time.sleep(5)
    return x

# does not block, returns Thread object
y = long_task(10)
print y

# this blocks, waiting for the result
result = y.result_queue.get()
print result

The decorated function creates a new thread each time it’s called and returns a Thread object that contains the queue that will receive the result.

UPDATE

It’s been quite a while since I posted this answer, but it still gets views so I thought I would update it to reflect the way I do this in newer versions of Python:

Python 3.2 added in the concurrent.futures module which provides a high-level interface for parallel tasks. It provides ThreadPoolExecutor and ProcessPoolExecutor, so you can use a thread or process pool with the same api.

One benefit of this api is that submitting a task to an Executor returns a Future object, which will complete with the return value of the callable you submit.

This makes attaching a queue object unnecessary, which simplifies the decorator quite a bit:

_DEFAULT_POOL = ThreadPoolExecutor()

def threadpool(f, executor=None):
    @wraps(f)
    def wrap(*args, **kwargs):
        return (executor or _DEFAULT_POOL).submit(f, *args, **kwargs)

    return wrap

This will use a default module threadpool executor if one is not passed in.

The usage is very similar to before:

@threadpool
def long_task(x):
    import time
    x = x + 5
    time.sleep(5)
    return x

# does not block, returns Future object
y = long_task(10)
print y

# this blocks, waiting for the result
result = y.result()
print result

If you’re using Python 3.4+, one really nice feature of using this method (and Future objects in general) is that the returned future can be wrapped to turn it into an asyncio.Future with asyncio.wrap_future. This makes it work easily with coroutines:

result = await asyncio.wrap_future(long_task(10))

If you don’t need access to the underlying concurrent.Future object, you can include the wrap in the decorator:

_DEFAULT_POOL = ThreadPoolExecutor()

def threadpool(f, executor=None):
    @wraps(f)
    def wrap(*args, **kwargs):
        return asyncio.wrap_future((executor or _DEFAULT_POOL).submit(f, *args, **kwargs))

    return wrap

Then, whenever you need to push cpu intensive or blocking code off the event loop thread, you can put it in a decorated function:

@threadpool
def some_long_calculation():
    ...

# this will suspend while the function is executed on a threadpool
result = await some_long_calculation()

The Answer 5

65 people think this answer is useful

Another solution that doesn’t require changing your existing code:

import Queue             # Python 2.x
#from queue import Queue # Python 3.x

from threading import Thread

def foo(bar):
    print 'hello {0}'.format(bar)
    return 'foo'

que = Queue.Queue()      # Python 2.x
#que = Queue()           # Python 3.x

t = Thread(target=lambda q, arg1: q.put(foo(arg1)), args=(que, 'world!'))
t.start()
t.join()
result = que.get()
print result

It can be also easily adjusted to a multi-threaded environment:

import Queue             # Python 2.x
#from queue import Queue # Python 3.x
from threading import Thread

def foo(bar):
    print 'hello {0}'.format(bar)
    return 'foo'

que = Queue.Queue()      # Python 2.x
#que = Queue()           # Python 3.x

threads_list = list()

t = Thread(target=lambda q, arg1: q.put(foo(arg1)), args=(que, 'world!'))
t.start()
threads_list.append(t)

# Add more threads here
...
threads_list.append(t2)
...
threads_list.append(t3)
...

# Join all the threads
for t in threads_list:
    t.join()

# Check thread's return value
while not que.empty():
    result = que.get()
    print result

The Answer 6

26 people think this answer is useful

Parris / kindall’s answer join/return answer ported to Python 3:

from threading import Thread

def foo(bar):
    print('hello {0}'.format(bar))
    return "foo"

class ThreadWithReturnValue(Thread):
    def __init__(self, group=None, target=None, name=None, args=(), kwargs=None, *, daemon=None):
        Thread.__init__(self, group, target, name, args, kwargs, daemon=daemon)

        self._return = None

    def run(self):
        if self._target is not None:
            self._return = self._target(*self._args, **self._kwargs)

    def join(self):
        Thread.join(self)
        return self._return


twrv = ThreadWithReturnValue(target=foo, args=('world!',))

twrv.start()
print(twrv.join())   # prints foo

Note, the Thread class is implemented differently in Python 3.

The Answer 7

24 people think this answer is useful

I stole kindall’s answer and cleaned it up just a little bit.

The key part is adding *args and **kwargs to join() in order to handle the timeout

class threadWithReturn(Thread):
    def __init__(self, *args, **kwargs):
        super(threadWithReturn, self).__init__(*args, **kwargs)

        self._return = None

    def run(self):
        if self._Thread__target is not None:
            self._return = self._Thread__target(*self._Thread__args, **self._Thread__kwargs)

    def join(self, *args, **kwargs):
        super(threadWithReturn, self).join(*args, **kwargs)

        return self._return

UPDATED ANSWER BELOW

This is my most popularly upvoted answer, so I decided to update with code that will run on both py2 and py3.

Additionally, I see many answers to this question that show a lack of comprehension regarding Thread.join(). Some completely fail to handle the timeout arg. But there is also a corner-case that you should be aware of regarding instances when you have (1) a target function that can return None and (2) you also pass the timeout arg to join(). Please see “TEST 4” to understand this corner case.

ThreadWithReturn class that works with py2 and py3:

import sys
from threading import Thread
from builtins import super    # https://stackoverflow.com/a/30159479

if sys.version_info >= (3, 0):
    _thread_target_key = '_target'
    _thread_args_key = '_args'
    _thread_kwargs_key = '_kwargs'
else:
    _thread_target_key = '_Thread__target'
    _thread_args_key = '_Thread__args'
    _thread_kwargs_key = '_Thread__kwargs'

class ThreadWithReturn(Thread):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self._return = None

    def run(self):
        target = getattr(self, _thread_target_key)
        if not target is None:
            self._return = target(
                *getattr(self, _thread_args_key),
                **getattr(self, _thread_kwargs_key)
            )

    def join(self, *args, **kwargs):
        super().join(*args, **kwargs)
        return self._return

Some sample tests are shown below:

import time, random

# TEST TARGET FUNCTION
def giveMe(arg, seconds=None):
    if not seconds is None:
        time.sleep(seconds)
    return arg

# TEST 1
my_thread = ThreadWithReturn(target=giveMe, args=('stringy',))
my_thread.start()
returned = my_thread.join()
# (returned == 'stringy')

# TEST 2
my_thread = ThreadWithReturn(target=giveMe, args=(None,))
my_thread.start()
returned = my_thread.join()
# (returned is None)

# TEST 3
my_thread = ThreadWithReturn(target=giveMe, args=('stringy',), kwargs={'seconds': 5})
my_thread.start()
returned = my_thread.join(timeout=2)
# (returned is None) # because join() timed out before giveMe() finished

# TEST 4
my_thread = ThreadWithReturn(target=giveMe, args=(None,), kwargs={'seconds': 5})
my_thread.start()
returned = my_thread.join(timeout=random.randint(1, 10))

Can you identify the corner-case that we may possibly encounter with TEST 4?

The problem is that we expect giveMe() to return None (see TEST 2), but we also expect join() to return None if it times out.

returned is None means either:

(1) that’s what giveMe() returned, or

(2) join() timed out

This example is trivial since we know that giveMe() will always return None. But in real-world instance (where the target may legitimately return None or something else) we’d want to explicitly check for what happened.

Below is how to address this corner-case:

# TEST 4
my_thread = ThreadWithReturn(target=giveMe, args=(None,), kwargs={'seconds': 5})
my_thread.start()
returned = my_thread.join(timeout=random.randint(1, 10))

if my_thread.isAlive():
    # returned is None because join() timed out
    # this also means that giveMe() is still running in the background
    pass
    # handle this based on your app's logic
else:
    # join() is finished, and so is giveMe()
    # BUT we could also be in a race condition, so we need to update returned, just in case
    returned = my_thread.join()

The Answer 8

16 people think this answer is useful

Using Queue :

import threading, queue

def calc_square(num, out_queue1):
  l = []
  for x in num:
    l.append(x*x)
  out_queue1.put(l)


arr = [1,2,3,4,5,6,7,8,9,10]
out_queue1=queue.Queue()
t1=threading.Thread(target=calc_square, args=(arr,out_queue1))
t1.start()
t1.join()
print (out_queue1.get())

The Answer 9

6 people think this answer is useful

My solution to the problem is to wrap the function and thread in a class. Does not require using pools,queues, or c type variable passing. It is also non blocking. You check status instead. See example of how to use it at end of code.

import threading

class ThreadWorker():
    '''
    The basic idea is given a function create an object.
    The object can then run the function in a thread.
    It provides a wrapper to start it,check its status,and get data out the function.
    '''
    def __init__(self,func):
        self.thread = None
        self.data = None
        self.func = self.save_data(func)

    def save_data(self,func):
        '''modify function to save its returned data'''
        def new_func(*args, **kwargs):
            self.data=func(*args, **kwargs)

        return new_func

    def start(self,params):
        self.data = None
        if self.thread is not None:
            if self.thread.isAlive():
                return 'running' #could raise exception here

        #unless thread exists and is alive start or restart it
        self.thread = threading.Thread(target=self.func,args=params)
        self.thread.start()
        return 'started'

    def status(self):
        if self.thread is None:
            return 'not_started'
        else:
            if self.thread.isAlive():
                return 'running'
            else:
                return 'finished'

    def get_results(self):
        if self.thread is None:
            return 'not_started' #could return exception
        else:
            if self.thread.isAlive():
                return 'running'
            else:
                return self.data

def add(x,y):
    return x +y

add_worker = ThreadWorker(add)
print add_worker.start((1,2,))
print add_worker.status()
print add_worker.get_results()

The Answer 10

4 people think this answer is useful

I’m using this wrapper, which comfortably turns any function for running in a Thread – taking care of its return value or exception. It doesn’t add Queue overhead.

def threading_func(f):
    """Decorator for running a function in a thread and handling its return
    value or exception"""
    def start(*args, **kw):
        def run():
            try:
                th.ret = f(*args, **kw)
            except:
                th.exc = sys.exc_info()
        def get(timeout=None):
            th.join(timeout)
            if th.exc:
                raise th.exc[0], th.exc[1], th.exc[2] # py2
                ##raise th.exc[1] #py3                
            return th.ret
        th = threading.Thread(None, run)
        th.exc = None
        th.get = get
        th.start()
        return th
    return start

Usage Examples

def f(x):
    return 2.5 * x
th = threading_func(f)(4)
print("still running?:", th.is_alive())
print("result:", th.get(timeout=1.0))

@threading_func
def th_mul(a, b):
    return a * b
th = th_mul("text", 2.5)

try:
    print(th.get())
except TypeError:
    print("exception thrown ok.")

Notes on threading module

Comfortable return value & exception handling of a threaded function is a frequent “Pythonic” need and should indeed already be offered by the threading module – possibly directly in the standard Thread class. ThreadPool has way too much overhead for simple tasks – 3 managing threads, lots of bureaucracy. Unfortunately Thread‘s layout was copied from Java originally – which you see e.g. from the still useless 1st (!) constructor parameter group.

The Answer 11

4 people think this answer is useful

Taking into consideration @iman comment on @JakeBiesinger answer I have recomposed it to have various number of threads:

from multiprocessing.pool import ThreadPool

def foo(bar, baz):
    print 'hello {0}'.format(bar)
    return 'foo' + baz

numOfThreads = 3 
results = []

pool = ThreadPool(numOfThreads)

for i in range(0, numOfThreads):
    results.append(pool.apply_async(foo, ('world', 'foo'))) # tuple of args for foo)

# do some other stuff in the main process
# ...
# ...

results = [r.get() for r in results]
print results

pool.close()
pool.join()

Cheers,

Guy.

The Answer 12

3 people think this answer is useful

join always return None, i think you should subclass Thread to handle return codes and so.

The Answer 13

2 people think this answer is useful

You can define a mutable above the scope of the threaded function, and add the result to that. (I also modified the code to be python3 compatible)

returns = {}
def foo(bar):
    print('hello {0}'.format(bar))
    returns[bar] = 'foo'

from threading import Thread
t = Thread(target=foo, args=('world!',))
t.start()
t.join()
print(returns)

This returns {'world!': 'foo'}

If you use the function input as the key to your results dict, every unique input is guaranteed to give an entry in the results

The Answer 14

2 people think this answer is useful

GuySoft’s idea is great, but I think the object does not necessarily have to inherit from Thread and start() could be removed from interface:

from threading import Thread
import queue
class ThreadWithReturnValue(object):
    def __init__(self, target=None, args=(), **kwargs):
        self._que = queue.Queue()
        self._t = Thread(target=lambda q,arg1,kwargs1: q.put(target(*arg1, **kwargs1)) ,
                args=(self._que, args, kwargs), )
        self._t.start()

    def join(self):
        self._t.join()
        return self._que.get()


def foo(bar):
    print('hello {0}'.format(bar))
    return "foo"

twrv = ThreadWithReturnValue(target=foo, args=('world!',))

print(twrv.join())   # prints foo

The Answer 15

1 people think this answer is useful

Define your target to
1) take an argument q
2) replace any statements return foo with q.put(foo); return

so a function

def func(a):
    ans = a * a
    return ans

would become

def func(a, q):
    ans = a * a
    q.put(ans)
    return

and then you would proceed as such

from Queue import Queue
from threading import Thread

ans_q = Queue()
arg_tups = [(i, ans_q) for i in xrange(10)]

threads = [Thread(target=func, args=arg_tup) for arg_tup in arg_tups]
_ = [t.start() for t in threads]
_ = [t.join() for t in threads]
results = [q.get() for _ in xrange(len(threads))]

And you can use function decorators/wrappers to make it so you can use your existing functions as target without modifying them, but follow this basic scheme.

The Answer 16

1 people think this answer is useful

As mentioned multiprocessing pool is much slower than basic threading. Using queues as proposeded in some answers here is a very effective alternative. I have use it with dictionaries in order to be able run a lot of small threads and recuperate multiple answers by combining them with dictionaries:

#!/usr/bin/env python3

import threading
# use Queue for python2
import queue
import random

LETTERS = 'abcdefghijklmnopqrstuvwxyz'
LETTERS = [ x for x in LETTERS ]

NUMBERS = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

def randoms(k, q):
    result = dict()
    result['letter'] = random.choice(LETTERS)
    result['number'] = random.choice(NUMBERS)
    q.put({k: result})

threads = list()
q = queue.Queue()
results = dict()

for name in ('alpha', 'oscar', 'yankee',):
    threads.append( threading.Thread(target=randoms, args=(name, q)) )
    threads[-1].start()
_ = [ t.join() for t in threads ]
while not q.empty():
    results.update(q.get())

print(results)

The Answer 17

1 people think this answer is useful

Based of what kindall mentioned, here’s the more generic solution that works with Python3.

import threading

class ThreadWithReturnValue(threading.Thread):
    def __init__(self, *init_args, **init_kwargs):
        threading.Thread.__init__(self, *init_args, **init_kwargs)
        self._return = None
    def run(self):
        self._return = self._target(*self._args, **self._kwargs)
    def join(self):
        threading.Thread.join(self)
        return self._return

Usage

        th = ThreadWithReturnValue(target=requests.get, args=('http://www.google.com',))
        th.start()
        response = th.join()
        response.status_code  # => 200

The Answer 18

0 people think this answer is useful

One usual solution is to wrap your function foo with a decorator like

result = queue.Queue()

def task_wrapper(*args):
    result.put(target(*args))

Then the whole code may looks like that

result = queue.Queue()

def task_wrapper(*args):
    result.put(target(*args))

threads = [threading.Thread(target=task_wrapper, args=args) for args in args_list]

for t in threads:
    t.start()
    while(True):
        if(len(threading.enumerate()) < max_num):
            break
for t in threads:
    t.join()
return result

Note

One important issue is that the return values may be unorderred. (In fact, the return value is not necessarily saved to the queue, since you can choose arbitrary thread-safe data structure )

The Answer 19

0 people think this answer is useful

Kindall’s answer in Python3

class ThreadWithReturnValue(Thread):
    def __init__(self, group=None, target=None, name=None,
                 args=(), kwargs={}, *, daemon=None):
        Thread.__init__(self, group, target, name, args, kwargs, daemon)
        self._return = None 

    def run(self):
        try:
            if self._target:
                self._return = self._target(*self._args, **self._kwargs)
        finally:
            del self._target, self._args, self._kwargs 

    def join(self,timeout=None):
        Thread.join(self,timeout)
        return self._return

The Answer 20

0 people think this answer is useful

Here is the version that I created of @Kindall’s answer https://stackoverflow.com/a/6894023/12900787

This version makes it so that all you have to do is input your command with arguments to create the new thread.

(I have also included a couple of tests)

this was made with python 3.8


from threading import Thread
from typing import Any


#def threader(com):  # my original Version (Ignore this)
#    try:
#        threader = Thread(target = com)
#        threader.start()
#    except Exception as e:
#        print(e)
#        print('Could not start thread')


def test(plug, plug2, plug3):
    print(f"hello {plug}")
    print(f'I am the second plug : {plug2}')
    print(plug3)
    return 'I am the return Value!'


def test2(msg):
    return f'I am from the second test: {msg}'


def test3():
    print('hello world')


def NewThread(com, Returning: bool, *arguments) -> Any:
    """
    Will create a new thread for a function/command.

    :param com: Command to be Executed
    :param arguments: Arguments to be sent to Command
    :param Returning: True/False Will this command need to return anything
    """
    
    class NewThreadWorker(Thread):
        def __init__(self, group = None, target = None, name = None, args = (), kwargs = None, *,
                     daemon = None):
            Thread.__init__(self, group, target, name, args, kwargs, daemon = daemon)
            
            self._return = None
        
        def run(self):
            if self._target is not None:
                self._return = self._target(*self._args, **self._kwargs)
        
        def join(self):
            Thread.join(self)
            return self._return
    
    ntw = NewThreadWorker(target = com, args = (*arguments,))
    ntw.start()
    if Returning:
        return ntw.join()


if __name__ == "__main__":
    # threader(test('world'))
    print(NewThread(test, True, 'hi', 'test', test2('hi')))
    NewThread(test3, True)


Hope that this is useful to someone 🙂

The Answer 21

0 people think this answer is useful

Most answers I’ve found are long and require being familiar with other modules or advanced python features, and will be rather confusing to someone unless they’re already familiar with everything the answer talks about.

Working code for a simplified approach:

import threading, time, random

class ThreadWithResult(threading.Thread):
    def __init__(self, target, args):
        self.target = target
        self.args = args
        def function():
            self.result = self.target(*self.args)
        super().__init__(target=function, args=())

def function_to_thread(n):
    count = 0
    while count < 3:
            print(f'still running thread {n}')
            count +=1
            time.sleep(3)
    result = random.random()
    print(f'Return value of thread {n} should be: {result}')
    return result


def main():
    thread1 = ThreadWithResult(target=function_to_thread, args=(1,))
    thread2 = ThreadWithResult(target=function_to_thread, args=(2,))
    thread1.start()
    thread2.start()
    thread1.join()
    thread2.join()
    print(thread1.result)
    print(thread2.result)

main()

Explanation: I wanted to simplify things significantly, so I created a ThreadWithResult class and had it inherit from threading.Thread. The nested function function in __init__ calls the threaded function we want to save the value of, and saves the result of that as the instance attribute self.result after the thread finishes executing.

Creating an instance of this is identical to creating an instance of threading.Thread. Pass in the function you want to run on a new thread to the target argument and any arguments that your function might need to the args argument. (I realize there are other arguments you can include, but omitted them for simplicity since this suffices for typical use cases.)

e.g.

my_thread = ThreadWithResult(target=my_function, args=(arg1, arg2, arg3))

I think this is significantly easier to understand than the vast majority of answers, and this approach requires no extra imports! I included the time and random module to simulate the behavior of a thread, but they’re not required to achieve the functionality asked in the original question.

I know I’m answering this looong after the question was asked, but I hope this can help more people in the future!

The Answer 22

-1 people think this answer is useful

I know this thread is old…. but I faced the same problem… If you are willing to use thread.join()

import threading

class test:

    def __init__(self):
        self.msg=""

    def hello(self,bar):
        print('hello {}'.format(bar))
        self.msg="foo"


    def main(self):
        thread = threading.Thread(target=self.hello, args=('world!',))
        thread.start()
        thread.join()
        print(self.msg)

g=test()
g.main()

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