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 
//Product of three numbers
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class ProductOfThreeNumbers
{
	public static void main(String args[]) throws IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		int t = Integer.parseInt(br.readLine());
		while(t-- > 0){
			int n = Integer.parseInt(br.readLine());
			boolean flag = false;
			for(int i = 2; i < Math.floor(n/i); ++i){
				if(n/i == (double)n/i){
					int[] a = factors(n/i, i);
					if(a.length == 2){
						out.println("YES");
						out.println(i+" "+a[0]+" "+a[1]);
						flag = true;
						break;
					}
				}
			}
			if(!flag)
				out.println("NO");
			out.flush();
		}
	}
	static int[] factors(int n, int a){
		int end = (int)Math.sqrt(n);
		for(int i = a+1; i <= end; ++i){
			if(n%i == 0 && i != a){
				if(i != n/i){
					return new int[]{i, n/i};
				}
			}
		}
		return new int[]{0};
	}
}