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 int maxn = 1010;
int n;
int num[maxn], cnt;
int seq[maxn];
bool Find(int res, int now) {
  for (int i = 1; i <= now; i++) {
    if (seq[i] == res) return true;
  }
  return false;
}
void Init();
int main() {
  int T;
  scanf("%d", &T);
  while (T--) Init();
  return 0;
}
void Init() {
  scanf("%d", &n);
  int cun = n;
  cnt = 0;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) {
        n /= i;
        num[++cnt] = i;
      }
    }
  }
  if (n > 1) num[++cnt] = n;
  int now = 0, res = 1, i;
  for (i = 1; i <= cnt; i++) {
    res *= num[i];
    if (!Find(res, now)) {
      seq[++now] = res;
      seq[now] = res;
      if (now == 3) {
        seq[now] = cun / seq[1] / seq[2];
        break;
      }
      res = 1;
    }
  }
  if (now == 3) {
    puts("YES");
    for (int i = 1; i <= 3; i++) printf("%d ", seq[i]);
    puts("");
  } else
    puts("NO");
}