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;
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<T> &v) {
  for (typename vector<T>::size_type i = 0; i < v.size(); ++i)
    out << v[i] << " ";
  out << "\n";
  return out;
}
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<pair<T, T> > &v) {
  for (size_t i = 0; i < v.size(); ++i)
    out << "(" << v[i].first << ", " << v[i].second << ") ";
  out << "\n";
  return out;
}
template <typename T>
std::ostream &operator<<(std::ostream &out, vector<vector<T> > &v) {
  for (size_t i = 0; i < v.size(); ++i) {
    for (size_t j = 0; j < v[i].size(); ++j) {
      out << v[i][j] << " ";
    }
    out << "\n";
  }
  return out;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a = 1;
    for (int i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        a = i;
        break;
      }
    }
    if (a == 1) {
      cout << "NO\n";
      continue;
    }
    int b = 1;
    for (int i = a + 1; i * i <= n / a; ++i) {
      if ((n / a) % i == 0) {
        b = i;
        break;
      }
    }
    if (b == 1) {
      cout << "NO\n";
      continue;
    }
    int c = n / (a * b);
    if (c && c != 1 && c != a && c != b) {
      cout << "YES\n";
      cout << a << " " << b << " " << c << "\n";
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}