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 
from sys import stdin, stdout
from collections import defaultdict
import math
rl = lambda: stdin.readline()
rll = lambda: stdin.readline().split()

def main():
	cases = int(input())
	for line in stdin:
		n = int(line)
		ans = []
		f1 = 2 
		while f1 <= math.sqrt(n):
			if n % f1 == 0:
				ans.append(f1)
				break
			f1 += 1
		if len(ans) == 0:
			stdout.write("NO\n")
			continue
		m = n//f1
		f2 = f1 + 1
		while f2 <= math.sqrt(m):
			if m % f2 == 0:
				ans.append(f2)
				break
			f2 += 1
		if len(ans) == 1:
			stdout.write("NO\n")
			continue
		f3 = n//(f1*f2)
		if f3 not in {f1, f2}:
			stdout.write("YES\n")
			stdout.write(" ".join((str(x) for x in [f1, f2, f3])))
			stdout.write("\n")
		else:
			stdout.write("NO\n")
			
if __name__ == "__main__":
	main()