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;
void solve(long long N) {
  vector<pair<long long, long long>> factors;
  long long n = N;
  for (long long i = 2; i * i <= n; i++) {
    long long cnt = 0;
    if (n % i == 0) {
      while (n % i == 0) {
        cnt++;
        n /= i;
      }
      factors.push_back({i, cnt});
    }
    if (factors.size() == 3) break;
  }
  if (n != 1) {
    factors.push_back({n, 1});
  }
  if (factors.size() >= 2) {
    long long a = factors[0].first;
    long long b = factors[1].first;
    long long c = N / (a * b);
    if (c != a and c != b and c > 1) {
      cout << "YES" << '\n';
      cout << a << " " << b << " " << c << '\n';
    } else
      cout << "NO" << '\n';
  } else if (factors.size() == 1 and factors[0].second >= 6) {
    cout << "YES" << '\n';
    long long a = factors[0].first;
    long long b = a * a;
    long long c = N / (a * b);
    cout << a << " " << b << " " << c << '\n';
  } else
    cout << "NO" << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    solve(n);
  }
  return 0;
}