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;
long long q, i, j, n, a, b, sq, rez, x, rz[100];
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  long long r = a % b;
  while (r) {
    a = b;
    b = r;
    r = a % b;
  }
  return b;
}
int main() {
  rz[2] = 1;
  rz[3] = 1;
  rz[4] = 5;
  rz[5] = 1;
  rz[6] = 21;
  rz[7] = 1;
  rz[8] = 85;
  rz[9] = 73;
  rz[10] = 341;
  rz[11] = 89;
  rz[12] = 1365;
  rz[13] = 1;
  rz[14] = 5461;
  rz[15] = 4681;
  rz[16] = 21845;
  rz[17] = 1;
  rz[18] = 87381;
  rz[19] = 1;
  rz[20] = 349525;
  rz[21] = 299593;
  rz[22] = 1398101;
  rz[23] = 178481;
  rz[24] = 5592405;
  rz[25] = 1082401;
  cin >> q;
  while (q--) {
    cin >> n;
    j = n;
    x = n;
    while (x & (x - 1)) x &= x - 1;
    a = x;
    x <<= 1;
    x--;
    if (n != x)
      cout << x << "\n";
    else {
      x++;
      x /= 2;
      n = 0;
      while (x) n++, x /= 2;
      cout << rz[n] << "\n";
    }
  }
}
