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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Product_of_Three_Numbers {

	static int mod = (int) (1e9 + 7);

	public static void main(String[] args) throws java.lang.Exception {
		
/*
	Let a be the minimum divisor of n greater than 1. 
	Then let b be the minimum divisor of na that isn't equal a and 1. 
	If n/ab isn't equal a, b and 1 then the answer is "YES", 
	otherwise the answer is "NO".
	
	Time complexity: O(√n) per query.

*/
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		while (t-- > 0) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			int n=Integer.parseInt(st.nextToken());
			Set<Integer> h=new HashSet<>();
			for(int i=2; i*i<=n; i++) {
				if(n%i==0 && !h.contains(i)) {
					h.add(i);
					n/=i;
					break;
				}
			}
			for(int i=2; i*i<=n; i++) {
				if(n%i==0 && !h.contains(i)) {
					h.add(i);
					n/=i;
					break;
				}
			}
			if(h.size()<2 ||h.contains(n)|| n==1) System.out.println("NO");
			else {
				System.out.println("YES");
				h.add(n);
				for(int x: h) System.out.print(x+" ");
				System.out.println();
			}
		}
	}
}