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;
bool test(void) {
  long long N;
  scanf("%lld", &N);
  long long u = N;
  vector<pair<long long, int>> A;
  for (long long i = 2; i * i <= u; ++i) {
    int q = 0;
    while (u % i == 0) {
      ++q;
      u /= i;
    }
    if (q) {
      A.push_back({i, q});
    }
  }
  if (u > 1) A.push_back({u, 1});
  vector<long long> D{1};
  for (auto [p, q] : A) {
    int s = D.size();
    for (int i = 0; i < s; ++i) {
      long long u = D[i];
      for (int k = 0; k < q; ++k) {
        u *= p;
        D.push_back(u);
      }
    }
  }
  int s = D.size();
  for (int i = 0; i < s; ++i) {
    long long a = D[i];
    if (a == 1) continue;
    for (int j = 0; j < i - 1; ++j) {
      long long b = D[j];
      if (b == 1) continue;
      if (N % (a * b) == 0) {
        long long c = N / (a * b);
        if (c == 1) continue;
        if (a != c && b != c) {
          printf("YES\n");
          printf("%lld %lld %lld\n", a, b, c);
          return true;
        }
      }
    }
  }
  return false;
}
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) {
    if (!test()) printf("NO\n");
  }
  return 0;
}