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 
import math
t = int(input())

def primefactorize(n):
	prime_factors_list = []

	while n%2 == 0:
		prime_factors_list.append(2)
		n = n/2

	for i in range(3,int(math.sqrt(n))+1,2):

		while n%i == 0:
			prime_factors_list.append(i)
			n = n/i

	if n > 2:
		prime_factors_list.append(n)

	return prime_factors_list
def generate_factors(prime_factors_list,n):
	a = 1
	b = 1
	c = 1

	for i in range(len(prime_factors_list)):
		if a == 1 :
			a *= prime_factors_list[i]
		elif b == 1 or a == b:
			b *= prime_factors_list[i]
		else:
			c *= prime_factors_list[i]
	return a,b,c


for tt in range(t):
	n = int(input())
	prime_factors = primefactorize(n)
	# print(prime_factors)
	a,b,c = generate_factors(prime_factors,n)
	if (a!=1 and b!=1 and c!=1) and (a!=b and b!=c and a!=c):
		print("YES")
		print(str(int(a))+ " " + str(int(b)) + " " + str(int(c)))
	else:
		print("NO")