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;
const long double PI = acos(-1.0L);
const long double EPS = 1e-12L;
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
namespace task {
int t, n, k;
vector<int> p;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    k = n;
    p = {};
    for (int i = 2; i < min(n, 100000); ++i) {
      if (n % i == 0) {
        p.push_back(i);
        n /= i;
      }
    }
    if (p.size() < 2) {
      cout << "NO\n";
    } else if (p.size() == 2) {
      if (n <= p.back()) {
        cout << "NO\n";
      } else {
        cout << "YES\n" << p[0] << ' ' << p[1] << ' ' << n << '\n';
      }
    } else {
      cout << "YES\n" << p[0] << ' ' << p[1] << ' ' << k / p[0] / p[1] << '\n';
    }
  }
  return 0;
}
}  // namespace task
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(11);
  cout.setf(ios::fixed);
  return task::main();
}