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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const int maxn = 1e6 + 10;
const int inf = 2147483647;
int main() {
  vector<int> v, prim;
  vector<bool> isprim;
  int lim = sqrt(1.0 * 1000000000) + 10;
  isprim.resize(lim);
  isprim[0] = isprim[1] = 1;
  for (int i = 2; i <= lim; ++i) {
    if (isprim[i] == false) {
      for (int j = i + i; j <= lim; j = j + i) isprim[j] = true;
    }
  }
  for (int i = 2; i <= lim; ++i) {
    if (isprim[i] == false) prim.push_back(i);
  }
  int sz = prim.size();
  set<int> se;
  set<int>::iterator it;
  int T;
  scanf("%d", &T);
  while (T--) {
    int n = read();
    bool flag = false;
    se.clear();
    for (int i = 2; i * i <= n; ++i) {
      if (n % i == 0 && isprim[i] == false) {
        se.insert(i);
        n /= i;
        flag = true;
        break;
      }
    }
    if (!flag) {
      puts("NO");
      continue;
    }
    flag = false;
    for (int i = sqrt(n); i >= 2; i--) {
      if (n % i == 0 && se.count(i) == 0 && se.count(n / i) == 0 &&
          i != n / i) {
        flag = true;
        se.insert(i);
        se.insert(n / i);
        break;
      }
    }
    if (!flag) {
      puts("NO");
      continue;
    }
    puts("YES");
    for (it = se.begin(); it != se.end(); ++it) {
      printf("%d ", *it);
    }
    puts("");
  }
  return 0;
}