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 
from collections import defaultdict
def primeFactors(n): 
	d=defaultdict(int)

	while n % 2 == 0: 
		d[2]+=1
		n = n // 2

	for i in range(3,int(math.sqrt(n))+1,2): 
		
		while n % i== 0: 
			d[i]+=1
			n = n / i 
			
	if n > 2: 
		d[n]+=1 
	
	return d

t=int(input())
for i in range(t):
	n=int(input())
	d= primeFactors(n)
	# print(d)
	if len( d.keys() )>=3:
		print("YES")
		s=[]
		ww=1
		for j in list(d.keys())[:2]:
			s.append( int(j) )
			ww*=j**(d[j]-1)

		for j in list(d.keys())[2:]:
			ww*= int(j**d[j])
		
		s.append(ww)
		print(*s)
	elif len(list(d.keys()))==1:
		w,w1 = int(list(d.keys())[0]), int(d[list(d.keys())[0]]) 

		if w1>=6:
			print("YES")
			ans = "{} {} {}".format(w,w**2,w**(w1-3))
			print(ans)
		else:
			print("NO")
	elif len(list(d.keys()))==2:
		keys= list(map(int,list(d.keys())))
		value = list(map(int,list(d.values())))
		if sum(value)>=4:
			print("YES")
			ans = "{} {} {}".format( keys[0], keys[1], keys[0]**(value[0]-1) * keys[1]**(value[1]-1) )
			print(ans)
		else:
			print("NO")
