It's simple: To get past your check, just type password at the prompt :-) Try it!

What is the password?password
Access granted.

Explanation: input reads what you type, evaluates it as python code, and assigns the result to pass_true. So in the above, I just told it to assign the value of the variable password (which holds the real password) to pass_true.

But this code is broken even for regular use: If you type anything that doesn't look like a valid python expression, it'll trigger an error. Your example uses numbers as "passwords", but you couldn't use the same approach for arbitrary-text passwords: It wouldn't work at all, even with the correct password.

Another kind of problem is typing something damaging like os.remove(".profile") (or something worse and more elaborate) at the prompt. It won't log them in but you can see the damage it can do. Evaluating user input is just wrong, unless you're writing an interactive python interpreter.

Python 2 has raw_input() which just reads input. (In python 3, raw_input() has been renamed to input() and the old input() is gone).

Can you try the following:

test = int(input())
for __ in range(test):
    numberSize = input()
    li = input.split()

The first input is for total number of test cases, and we are iterating the total number of test cases. They have mentioned in in each test cases there are two inputs, which we are getting it inside the for loop.