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;
int gcd(int r, int p) {
  if (p == 0) return r;
  return gcd(p, r % p);
}
struct Test {
  int x, y, z;
};
long long int fact(long long int x) {
  long long int p = (fact(x - 1) * x) % 1000000007;
  return p;
}
void sieveOfEratosthenes(long long int N, long long int s[]) {
  vector<bool> prime(N + 1, false);
  for (long long int i = 2; i <= N; i += 2) s[i] = 2;
  for (long long int i = 3; i <= N; i += 2) {
    if (prime[i] == false) {
      s[i] = i;
      for (long long int j = i; j * i <= N; j += 2) {
        if (prime[i * j] == false) {
          prime[i * j] = true;
          s[i * j] = i;
        }
      }
    }
  }
}
map<long long int, long long int> poww;
vector<long long int> vec;
void generatePrimeFactors(long long int N) {
  long long int s[N + 1];
  sieveOfEratosthenes(N, s);
  long long int curr = s[N];
  int cnt = 1;
  while (N > 1) {
    N /= s[N];
    if (curr == s[N]) {
      cnt++;
      continue;
    }
    poww[curr] = cnt;
    vec.push_back(curr);
    if (vec.size() > 2) {
      return;
    }
    curr = s[N];
    cnt = 1;
  }
}
int main() {
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    int f = 0;
    cin >> n;
    long long int t, tt, ttt;
    for (int i = 2; i < sqrt(n); i++) {
      if (n % i == 0) {
        for (int j = 2; j < sqrt(n); j++) {
          if (n / i % j == 0 && i != j && n / (i * j) != j &&
              n / (i * j) != i) {
            f = 1;
            t = i;
            tt = j;
            ttt = n / (i * j);
            break;
          }
        }
      }
    }
    if (f == 1) {
      cout << "YES" << endl;
      cout << t << " " << tt << " " << ttt << endl;
    } else {
      cout << "NO" << endl;
    }
  }
}