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>
using namespace std;
int ans1, ans2, n, T;
inline int read() {
  int x = 0;
  int flag = 0;
  char ch;
  ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = 1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - 48;
    ch = getchar();
  }
  if (flag == 1) x = -x;
  return x;
}
inline bool solve(int x) {
  for (register int i = 2; i <= sqrt(x) + 1; i++) {
    if (x % i == 0 && i != x / i) {
      ans1 = i;
      ans2 = x / i;
      return true;
    }
  }
  return false;
}
int main() {
  T = read();
  while (T--) {
    n = read();
    int flag = 0;
    for (register int a = 2; a <= sqrt(n) + 1; a++) {
      if (n % a == 0) {
        int rest = n / a;
        for (register int i = 2; i <= sqrt(rest) + 1; i++) {
          if (rest % i == 0) {
            ans1 = i;
            ans2 = rest / i;
            if (ans1 != ans2 && ans1 != a && ans2 != a && ans1 != 1 &&
                ans2 != 1) {
              printf("YES\n%d %d %d\n", a, ans1, ans2);
              flag = 1;
              break;
            }
          }
        }
        if (flag == 1) break;
      }
    }
    if (!flag) {
      printf("NO\n");
    }
  }
}