Как узнать file_id на apkfiles.com?

r = requests.post('/submit-2', data=data)
file_id = r.url.split('=')[1]

fpath=open(r'D:\apk\Simple Gallery APK v 2.9.1.apk','rb')
                file2=fpath.read()               
                fpath.close()
                file={'apk': ('Simple Gallery APK v 2.9.1.apk', file2, 'application/vnd.android.package-archive', {'Expires': '0'})}
                paramfi= {'file_box': file,                                      
                        'button': 'Upload'
                                 }
                r = s.post(url_sub, headers=head, proxies=proxy, data=paramfi, files=file)
                #r = s.post(url_sub, headers=head, proxies=proxy, files=file)
                #r = s.post(url_sub, headers=head, proxies=proxy, data=paramfi)
                print(r.url)

https://www.apkfiles.com/submit?message=cannot%20find%20uploaded%20file,%20try%20again&message_type=error
cannot%20find%20uploaded%20file,%20try%20again&message_type
['https://www.apkfiles.com/submit-2']

import requests

r = requests.post('https://www.apkfiles.com/submit',
                  data={'button': 'Upload as guest'},
                  files={'file_box': open(r'/home/dmitriy/Desktop/example.apk' 
                         , 'rb')})
print(r.url)
# 'https://www.apkfiles.com/submit-2?file_id=592410ed84d83'
file_id = r.url.split('=')[1]
print(file_id)
# 592410ed84d83