Запуск EXE файла используя pynsist

Я смотрел пример здесь

[Application]
name = Lab2
version=1.0
entry_point = lab1:main

[Python]
version = 3.8.5

[Run]
SetOutPath "$APPS"
File "D:\PythonProj\KPI - Security data\lab1\lab2.exe"
Exec "$APPS\lab2.exe 'Regedit'"

, но я получаю ошибку

PS D:\PythonProj\KPI - Security data\lab1> pynsist .\installer.cfg
Traceback (most recent call last):
  File "c:\users\user\appdata\local\programs\python\python38-32\lib\runpy.py", line 194, in _run_module_as_main
    return _run_code(code, main_globals, None,
  File "c:\users\user\appdata\local\programs\python\python38-32\lib\runpy.py", line 87, in _run_code
    exec(code, run_globals)
  File "C:\Users\User\AppData\Local\Programs\Python\Python38-32\Scripts\pynsist.exe\__main__.py", line 7, in <module>
  File "c:\users\user\appdata\local\programs\python\python38-32\lib\site-packages\nsist\__init__.py", line 511, in main
    cfg = configreader.read_and_validate(config_file)
  File "c:\users\user\appdata\local\programs\python\python38-32\lib\site-packages\nsist\configreader.py", line 110, in read_and_validate
    if config.read(config_file) == []:
  File "c:\users\user\appdata\local\programs\python\python38-32\lib\configparser.py", line 697, in read
    self._read(fp, filename)
  File "c:\users\user\appdata\local\programs\python\python38-32\lib\configparser.py", line 1113, in _read
    raise e
configparser.ParsingError: Source contains parsing errors: 'installer.cfg'
        [line 10]: 'SetOutPath "$APPS"\n'
        [line 13]: 'Exec "$APPS\\lab2.exe \'Regedit\'"\n'