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.
#include <bits/stdc++.h>
using namespace std;
int pop(long long n) {
  int ans = 0;
  while (n) {
    ans++;
    n = n >> 1;
  }
  return ans;
}
int main() {
  long long i, j, k, m, n, t, p;
  cin >> t;
  long long bit[30];
  bit[1] = 1;
  bit[2] = 1;
  bit[3] = 1;
  bit[4] = 1;
  bit[5] = 5, bit[6] = 1, bit[7] = 21, bit[8] = 1, bit[9] = 85, bit[10] = 73,
  bit[11] = 341;
  bit[12] = 89, bit[13] = 1365, bit[14] = 1;
  bit[15] = 5461, bit[16] = 4681, bit[17] = 21845, bit[18] = 1, bit[19] = 87381,
  bit[20] = 1;
  bit[21] = 349525, bit[22] = 299593, bit[23] = 1398101;
  bit[24] = 178481, bit[25] = 5592405, bit[26] = 1082401;
  while (t--) {
    long long a;
    cin >> a;
    j = pop(a);
    k = pow(2, j) - 1;
    if (k == a) {
      cout << bit[j + 1] << endl;
    } else {
      cout << k << endl;
    }
  }
}
