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:
/*
javac c.java && java c
*/

import java.io.*;
import java.util.*;

public class c {
	public static void main(String[] args) { new c(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);

	int t;	
	int n;

	c() {
		t = in.nextInt();
		while (t-- > 0) {
			n = in.nextInt();
			int[] ans = new int[3];

			int lp = 2;
			boolean found = false;
			for (; lp * lp <= n; lp++) {
				if (n %lp == 0) {
					found = true;
					n /= lp;
					ans[0] = lp;
					break;
				}
			}
			if (!found) {
				out.println("NO");
			}
			else {
				int nlp = ans[0] + 1;
				found = false;
				for (; nlp * nlp <= n; nlp++)
					if (n %nlp == 0) {
						found = true;
						n /= nlp;
						ans[1] = nlp;
						break;
					}
				if (!found) out.println("NO");
				else {
					if (n < 2 || n == nlp) out.println("NO");
					else {
						out.println("YES");
						out.println(ans[0] + " " + ans[1] + " " + n);
					}
				}
			}
		}
		out.close();
	}

	
	
	int abs(int x) { if (x < 0) return -x; return x; }
	long abs(long x) { if (x < 0) return -x; return x; }
	int min(int a, int b) { if (a < b) return a; return b; }	
	int max(int a, int b) { if (a > b) return a; return b; }	
	long min(long a, long b) { if (a < b) return a; return b; }	
	long max(long a, long b) { if (a > b) return a; return b; }	
	int gcd(int a, int b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
	long gcd(long a, long b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
	long lcm(int a, int b) { return (((long) a) * b) / gcd(a, b); }
	long lcm(long a, long b) { return (a * b) / gcd(a, b); }

	void sort(int[] arr) {
		int sz = arr.length, j;
		Random r = new Random();
		for (int i = 0; i < sz; i++) {
			j = r.nextInt(i + 1);
			arr[i] = arr[j]^(arr[i]^(arr[j] = arr[i]));
		} Arrays.sort(arr);
	}

	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}
}