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 main() {
  int t, n, d, a, b, c, i;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    d = 2;
    vector<int> ord;
    while (d * d <= n) {
      while (n % d == 0) {
        ord.push_back(d);
        n /= d;
      }
      d++;
    }
    if (n > 1) ord.push_back(n);
    if (ord.size() < 3) {
      printf("NO\n");
      continue;
    }
    a = ord[0];
    b = ord[ord.size() - 1], i = ord.size() - 2;
    if (a == b && ord.size() > 2) b *= ord[ord.size() - 2], i = ord.size() - 3;
    c = 1;
    while (i > 0) c *= ord[i--];
    if (a == b || b == c || a == c || c == 1)
      printf("NO\n");
    else {
      printf("YES\n%d %d %d\n", a, b, c);
    }
  }
  return 0;
}