1110C · Meaningless Operations

Problem: Can the greatest common divisor and bitwise operations have anything in common? It is time to answer this question.

Suppose you are given a positive integer $a$. You want to choose some integer $b$ from $1$ to $a - 1$ inclusive in such a way that the greatest common divisor (GCD) of integers $a \oplus b$ and $a \> \& \> b$ is as large as possible. In other words, you'd like to compute the following function:

$f(a) = \max_{0 < b < a}{gcd(a \oplus b, a \> \& \> b)}.$

Here $\oplus$ denotes the bitwise XOR operation, and $\&$ denotes the bitwise AND operation.

The greatest common divisor of two integers $x$ and $y$ is the largest integer $g$ such that both $x$ and $y$ are divided by $g$ without remainder.

You are given $q$ integers $a_1, a_2, \ldots, a_q$. For each of these integers compute the largest possible value of the greatest common divisor (when $b$ is chosen optimally).

Input Format: The first line contains an integer $q$ ($1 \le q \le 10^3$) — the number of integers you need to compute the answer for.

After that $q$ integers are given, one per line: $a_1, a_2, \ldots, a_q$ ($2 \le a_i \le 2^{25} - 1$) — the integers you need to compute the answer for.

Output Format: For each integer, print the answer in the same order as the integers are given in input.

Note: For the first integer the optimal choice is $b = 1$, then $a \oplus b = 3$, $a \> \& \> b = 0$, and the greatest common divisor of $3$ and $0$ is $3$.

For the second integer one optimal choice is $b = 2$, then $a \oplus b = 1$, $a \> \& \> b = 2$, and the greatest common divisor of $1$ and $2$ is $1$.

For the third integer the optimal choice is $b = 2$, then $a \oplus b = 7$, $a \> \& \> b = 0$, and the greatest common divisor of $7$ and $0$ is $7$.