Problem: You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).

Output

For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.

Otherwise, print "YES" and any possible such representation.

Example

Input


5
64
32
97
2
12345


Output


YES
2 4 8 
NO
NO
NO
YES
3 5 823 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma warning(disable : 6031)
int main() {
  int c;
  scanf("%d", &c);
  int ans[3];
  while (c--) {
    memset(ans, 0, sizeof(ans));
    int n, now = -1;
    scanf("%d", &n);
    for (int t = 2; t <= sqrt(n); t++) {
      if (n % t == 0) {
        now++;
        ans[now] = t;
        if (now == 1) {
          now++;
          ans[now] = n / ans[0] / ans[1];
          if (ans[0] != ans[1] && ans[1] != ans[2] && ans[0] != ans[2] &&
              n == ans[0] * ans[1] * ans[2])
            break;
          else
            now = 0;
        }
      }
    }
    if (now == 2 && ans[0] != ans[1] && ans[1] != ans[2] && ans[0] != ans[2] &&
        n == ans[0] * ans[1] * ans[2])
      printf("YES\n%d %d %d\n", ans[0], ans[1], ans[2]);
    else
      printf("NO\n");
  }
  return 0;
}