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;
#pragma comment(linker, "/STACK:10240000000,10240000000")
char ch;
int bo;
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
inline void rd(int &x) {
  x = bo = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') bo = 1;
  for (; ch >= '0' && ch <= '9';
       x = (x << 1) + (x << 3) + ch - '0', ch = getchar())
    ;
  if (bo) x = -x;
}
inline void RD(long long &x) {
  x = bo = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') bo = 1;
  for (; ch >= '0' && ch <= '9';
       x = (x << 1) + (x << 3) + ch - '0', ch = getchar())
    ;
  if (bo) x = -x;
}
inline long long RD() {
  long long x;
  RD(x);
  return x;
}
inline int rd() {
  int x;
  rd(x);
  return x;
}
inline void RD(char *s) {
  for (ch = getchar(); blank(ch); ch = getchar())
    ;
  for (; !blank(ch); ch = getchar()) *s++ = ch;
  *s = 0;
}
inline void RD(char &c) {
  for (ch = getchar(); blank(c); ch = getchar())
    ;
}
template <class T>
inline void OT(T x) {
  static char buf[20];
  char *p1 = buf;
  if (!x) *p1++ = '0';
  if (x < 0) putchar('-'), x = -x;
  while (x) *p1++ = x % 10 + '0', x /= 10;
  while (p1-- != buf) putchar(*p1);
}
const int maxn = 1e3 + 10;
const int mod = 1e9 + 7;
vector<int> v;
int main() {
  for (int _ = rd(), Case = 1; Case <= _; Case++) {
    int n = rd();
    int rn = n;
    int num = 0;
    v.clear();
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        num++;
        n /= i;
        v.push_back(i);
        if (v.size() >= 3) break;
      }
    }
    if (v.size() < 3 && n != 1) v.push_back(n);
    if (v.size() < 3)
      puts("NO");
    else {
      int fl = 0;
      for (int i = 0; i < v.size(); i++)
        for (int j = i + 1; j < v.size(); j++)
          if (v[i] == v[j]) fl = 1;
      if (fl)
        puts("NO");
      else {
        puts("YES");
        printf("%d %d %d\n", v[0], v[1], rn / v[0] / v[1]);
      }
    }
  }
}