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)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers a ⊕ b and a \> \& \> b is as large as possible. In other words, you'd like to compute the following function:

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

Here ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR), and \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

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, …, a_q. For each of these integers compute the largest possible value of the greatest common divisor (when b is chosen optimally). 

Input

The first line contains an integer q (1 ≤ q ≤ 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, …, a_q (2 ≤ a_i ≤ 2^{25} - 1) — the integers you need to compute the answer for. 

Output

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

Example

Input


3
2
3
5


Output


3
1
7

Note

For the first integer the optimal choice is b = 1, then a ⊕ 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 ⊕ 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 ⊕ b = 7, a \> \& \> b = 0, and the greatest common divisor of 7 and 0 is 7.

import java.util.Scanner;

public class MeaninglessOperations {

    static int gcd(int a, int b) {
        int temp = 0;
        while (b != 0) {
            temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        int[] dp = new int[33554450];
        dp[3] = 1;
        dp[7] = 1;
        dp[15] = 5;
        dp[31] = 1;
        dp[63] = 21;
        dp[127] = 1;
        dp[255] = 85;
        dp[511] = 73;
        dp[1023] = 341;
        dp[2047] = 89;
        dp[4095] = 1365;
        dp[8191] = 1;
        dp[16383] = 5461;
        dp[32767] = 4681;
        dp[65535] = 21845;
        dp[131071] = 1;
        dp[262143] = 87381;
        dp[524287] = 1;
        dp[1048575] = 349525;
        dp[2097151] = 299593;
        dp[4194303] = 1398101;
        dp[8388607] = 178481;
        dp[16777215] = 5592405;
        dp[33554431] = 1082401;

        int q = scan.nextInt();
        while (q-- > 0) {
            int in = scan.nextInt();
            int lg = 0;
            while (Math.pow(2, lg) <= in) {
                ++lg;
            }
            if (Math.pow(2, lg) == in + 1) {
                System.out.println(dp[in]);
            } else {
                System.out.println((int) Math.pow(2, lg) - 1);
            }
        }

    }
}
