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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> f, c;
    int r = n;
    int co{0};
    while (r % 2 == 0) {
      r = r / 2;
      co++;
    }
    if (co > 0) {
      f.push_back(2);
      c.push_back(co);
    }
    for (int i = 3; i * i < n; i++) {
      co = 0;
      while (r % i == 0) {
        r = r / i;
        co++;
      }
      if (co > 0) {
        f.push_back(i);
        c.push_back(co);
      }
    }
    if (r > 2) {
      f.push_back(r);
      c.push_back(1);
    }
    if (f.size() >= 3) {
      cout << "YES\n";
      cout << f[0] << ' ' << f[1] << ' ';
      n = n / f[0];
      n = n / f[1];
      cout << n << '\n';
    } else if (f.size() == 2) {
      if (c[0] + c[1] >= 4) {
        cout << "YES\n";
        cout << f[0] << ' ' << f[1] << ' ';
        n = n / f[0];
        n = n / f[1];
        cout << n << '\n';
      } else {
        cout << "NO\n";
      }
    } else {
      if (c[0] > 5) {
        cout << "YES\n";
        cout << f[0] << ' ' << f[0] * f[0] << ' ';
        n = n / f[0];
        n = n / f[0];
        n = n / f[0];
        cout << n << '\n';
      } else {
        cout << "NO\n";
      }
    }
  }
  return 0;
}