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 GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long llinf = (1LL << 62);
const int inf = (1 << 30);
const int nmax = 1e3 + 50;
const int mod = 1e9 + 7;
using namespace std;
int n, x, t, i, a, b, c, bl, j;
vector<int> d;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    d.clear();
    for (i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        d.push_back(i);
        if (i * i != n) d.push_back(n / i);
      }
    }
    bl = 0;
    for (i = 0; i < (int)d.size(); i++) {
      for (j = i + 1; j < (int)d.size(); j++) {
        if (n % (d[i] * d[j]) == 0 && n / (d[i] * d[j]) != d[i] &&
            n / (d[i] * d[j]) != d[j] && n / (d[i] * d[j]) != 1) {
          a = d[i], b = d[j], c = n / (d[i] * d[j]);
          bl = 1;
          break;
        }
      }
    }
    if (!bl)
      cout << "NO\n";
    else
      cout << "YES\n" << a << " " << b << " " << c << '\n';
  }
  return 0;
}