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 <class T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
void setmax(T& a, T b) {
  a = max(a, b);
};
template <typename T>
void setmin(T& a, T b) {
  a = min(a, b);
};
int getPow(int a, int b) {
  int p = 1;
  for (int i = 0; i < b; i++) p *= a;
  return p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    map<int, int> fac;
    int n;
    cin >> n;
    int cpy = n;
    for (long long i = 2; i * i <= n; i++) {
      while (n % i == 0) {
        fac[i]++;
        n /= i;
      }
    }
    if (n > 1) {
      fac[n]++;
    }
    n = cpy;
    if (fac.size() == 1) {
      auto [p, e] = *fac.begin();
      if (e < 6) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        cout << p << " " << p * p << " " << (n / (p * p * p)) << '\n';
      }
    } else if (fac.size() == 2) {
      auto x = *fac.begin();
      auto y = *fac.rbegin();
      int p = x.first, e = x.second;
      int q = y.first, u = y.second;
      if (e > u) {
        swap(p, q);
        swap(e, u);
      }
      if ((e == 1 && u == 1) || (e == 1 && u == 2)) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        if (u == 2) {
          cout << q << " " << (q * p) << " " << p << '\n';
        } else {
          cout << q << " ";
          cout << getPow(q, u - 1) << " ";
          cout << getPow(p, e) << " ";
        }
      }
    } else {
      vector<int> vals(3, 1);
      int i = 0;
      for (auto [p, e] : fac) {
        vals[(i++) % 3] *= getPow(p, e);
      }
      cout << "YES\n";
      for (int x : vals) cout << x << " ";
      cout << '\n';
    }
  }
  return 0;
}