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_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, j, t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    if (n == 1) {
      cout << "NO" << endl;
      continue;
    }
    int m = n;
    map<int, int> mp;
    vector<int> v;
    for (i = 2; i <= sqrt(n) + 1 && m > 1; i++) {
      if (m % i == 0)
        v.push_back(i);
      else
        continue;
      while (m % i == 0) {
        m /= i;
        mp[i]++;
      }
    }
    if (m != 1) {
      v.push_back(m);
      mp[m]++;
    }
    if (v.size() == 0) {
      cout << "NO" << endl;
      continue;
    }
    if (v.size() > 2) {
      cout << "YES" << endl;
      cout << v[0] << " ";
      n /= v[0];
      cout << v[1] << " ";
      n /= v[1];
      cout << n << endl;
      continue;
    }
    if (v.size() == 1) {
      int x = mp[v[0]];
      if (x < 6) {
        cout << "NO" << endl;
        continue;
      }
      int y = v[0];
      n /= pow(y, 3);
      cout << "YES" << endl;
      cout << y << " " << y * y << " " << n << endl;
      continue;
    } else {
      int x = mp[v[0]], y = mp[v[1]];
      if ((x < 2 && y < 2) || x + y < 4) {
        cout << "NO" << endl;
        continue;
      }
      cout << "YES" << endl;
      n /= (v[0] * v[1]);
      cout << v[1] << " " << v[0] << " " << n << endl;
      continue;
    }
  }
  return 0;
}