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 fastIo() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
const long long N = 100000;
bitset<N + 5> p;
vector<long long> primes;
void spf() {
  primes.push_back(2);
  for (int i = 4; i <= N; i += 2) p[i] = 1;
  for (long long i = 3; i <= N; i += 2) {
    if (p[i] == 0) {
      primes.push_back(i);
      for (long long j = i * i; j <= N; j += i) p[j] = 1;
    }
  }
}
vector<long long> factorise(long long n) {
  vector<long long> ans;
  for (long long i = 0; primes[i] * primes[i] <= n; i++) {
    if (n % primes[i] == 0) {
      while (n % primes[i] == 0) {
        ans.push_back(primes[i]);
        n /= primes[i];
      }
    }
  }
  if (n != 1) ans.push_back(n);
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> ans = factorise(n);
  if (ans.size() < 3) {
    cout << "NO\n";
    return;
  }
  long long a = ans[0], b = ans[1], c = 1, s = 1;
  if (a == b && ans.size() >= 4)
    b *= ans[2], s = 2;
  else if (a == b) {
    cout << "NO\n";
    return;
  }
  for (int i = s + 1; i < ans.size(); i++) c *= ans[i];
  if (c == 1 || c == a || c == b) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n" << a << " " << b << " " << c << endl;
}
int main() {
  spf();
  fastIo();
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}