Cirno's Perfect Bitmasks Classroom
Even if it's a really easy question, she won't be able to answer it— Perfect Memento in Strict SenseCirno's perfect bitmasks classroom has just started!Cirno gave her students a positive integer $$$x$$$. As an assignment, her students need to find the minimum positive integer $$$y$$$, which satisfies the following two conditions:$$$$$$x\ \texttt{and}\ y > 0$$$$$$ $$$$$$x\ \texttt{xor}\ y > 0$$$$$$Where $$$\texttt{and}$$$ is the bitwise AND operation, and $$$\texttt{xor}$$$ is the bitwise XOR operation.Among the students was Mystia, who was truly baffled by all these new operators. Please help her!
Test case 1: $$$1\; \texttt{and}\; 3=1>0$$$, $$$1\; \texttt{xor}\; 3=2>0$$$.Test case 2: $$$2\; \texttt{and}\; 3=2>0$$$, $$$2\; \texttt{xor}\; 3=1>0$$$.
The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of input test cases.For each test case, the only line of input contains one integer $$$x$$$ ($$$1 \leq x \leq 2^{30}$$$).
For each test case, print a single integer — the minimum number of $$$y$$$.
t = int(input())
for _ in range(t):
    x = int(input())
    if x == 1:
        print(3)
    else:
        y = x & -x
        if y == x:
            print(x + 1)
        else:
            print(y)
t = int(input())
for _ in range(t):
    x = int(input())
    if x == 1:
        print(3)
    else:
        y = x & -x
        if y == x:
            print(x + 1)
        else:
            print(y)
for _ in range(int(input())):
    a = int(input())
    y = a & -a
    if y == a:
        if a == 1:
            y += 1
        y += 1
    print(y)