remove server name to get public launch

app.py

@@ -63,7 +63,7 @@ class WhisperModel:
     def transcribe(self, audio_path):
         return self.model.transcribe(audio_path, suppress_tokens=self.supress_tokens, word_timestamps=True)["segments"]
 
-
+@spaces.GPU(duration=120)
 class WhisperxModel:
     def __init__(self, model_name, align_model: WhisperxAlignModel):
         from whisperx import load_model
@@ -605,7 +605,6 @@ if __name__ == "__main__":
     parser.add_argument("--models-path", default="./pretrained_models", help="Path to voicecraft models directory")
     parser.add_argument("--port", default=7860, type=int, help="App port")
     parser.add_argument("--share", action="store_true", help="Launch with public url")
-    parser.add_argument("--server_name", default="127.0.0.1", type=str, help="Server name for launching the app. 127.0.0.1 for localhost; 0.0.0.0 to allow access from other machines in the local network. Might also give access to external users depends on the firewall settings.")
 
     os.environ["USER"] = os.getenv("USER", "user")
     args = parser.parse_args()
@@ -614,4 +613,4 @@ if __name__ == "__main__":
     MODELS_PATH = args.models_path
 
     app = get_app()
-    app.queue().launch(share=args.share, server_name=args.server_name, server_port=args.port)
+    app.queue().launch(share=args.share, server_port=args.port)