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 MOD = 1000000007;
const int MAX = 1000005;
const double PI = acos(-1.0);
int SetBit(int n, int X) { return n | (1 << X); }
int ClearBit(int n, int X) { return n & ~(1 << X); }
int ToggleBit(int n, int X) { return n ^ (1 << X); }
bool CheckBit(int n, int X) { return (bool)(n & (1 << X)); }
void doTheTask(int n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      int d = n / i;
      for (int j = i + 1; j * j < d; j++) {
        if (d % j == 0) {
          printf("YES\n%d %d %d\n", i, j, d / j);
          return;
        }
      }
    }
  }
  printf("NO\n");
}
int main(void) {
  int tc, n, i, j, a, b, c;
  scanf("%d", &tc);
  while (tc--) {
    scanf("%d", &n);
    doTheTask(n);
  }
  return 0;
}