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 pri[100003], top, n;
bool ntp[100002];
struct node {
  int a, b;
  bool operator<(const node &rhs) const { return b > rhs.b; }
};
void euler() {
  for (int i = 2; i <= 100000; ++i) {
    if (!ntp[i]) pri[top++] = i;
    for (int j = 0; j < top && pri[j] * i <= 100000; ++j) {
      ntp[pri[j] * i] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  euler();
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    int bn = n;
    vector<node> v;
    int sz = 1;
    for (int i = 0; i < top && n != 1; ++i) {
      if (n % pri[i] == 0) {
        node x;
        x.a = pri[i];
        x.b = 0;
        while (n % pri[i] == 0) ++x.b, n /= pri[i];
        sz *= (x.b + 1);
        v.push_back(x);
      }
    }
    if (n != 1) {
      node x;
      x.a = n;
      x.b = 1;
      v.push_back(x);
    }
    if (v.size() == 0) {
      cout << "NO\n";
    } else if (v.size() == 1) {
      if (v[0].b < 6) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        const int &t = v[0].a;
        cout << t << ' ' << t * t << ' ' << bn / (t * t * t) << '\n';
      }
    } else if (v.size() == 2) {
      if (v[0].b + v[1].b < 4) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        cout << v[0].a << ' ' << v[1].a << ' ' << bn / v[0].a / v[1].a << '\n';
      }
    } else {
      cout << "YES\n";
      cout << v[0].a << ' ' << v[1].a << ' ' << bn / v[0].a / v[1].a << '\n';
    }
  }
  return 0;
}