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;
long long gcd(long long a, long long b);
long long nCr(long long n, long long r);
long long pow(long long b, long long n);
void read(long long a[], long long n);
void solve() {
  long long n;
  cin >> n;
  vector<long long> v;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      v.push_back(i);
      n /= i;
      if (v.size() == 2) {
        v.push_back(n);
        break;
      }
    }
  }
  if (v.size() != 3) {
    cout << "NO" << endl;
    return;
  }
  if (v[0] != v[1] && v[0] != v[1] && v[1] != v[2]) {
    cout << "YES" << endl;
    cout << v[0] << " " << v[1] << " " << v[2] << endl;
  } else {
    cout << "NO" << endl;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long nCr(long long n, long long r) {
  if (r > n) return 0;
  if (r * 2 > n) r = n - r;
  if (r == 0) return 1;
  long long ans = n;
  for (long long i = 2; i <= r; i++) {
    ans *= (n - i + 1);
    ans /= i;
  }
  return ans;
}
long long pow(long long b, long long n) {
  long long p = b, ans = 1;
  while (n) {
    if (n & 1) {
      ans *= p;
    }
    p *= p;
    n = n >> 1;
  }
  return ans;
}
void read(long long a[], long long n) {
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
}