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 
"""
NTC here
"""
import sys
inp= sys.stdin.readline
input = lambda : inp().strip()
flush= sys.stdout.flush
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(2**25)

def iin(): return int(input())
def lin(): return list(map(int, input().split()))

# range = xrange
# input = raw_input
def factors(a):
    fact = []
    if a % 2 == 0:
        ch = 0
        while a % 2 == 0:
            ch += 1
            a //= 2
        fact.extend([2]*ch)
    i = 3
    while i*i <= a:
        if a % i == 0:
            ch = 0
            while a % i == 0:
                ch += 1
                a //= i
            fact.extend([i]*ch)
        i += 2
    if a > 1:
        fact.append(a)
    return fact

def main():
    t = iin()
    ans = ['YES', "NO"]
    while t:
        t-=1
        n = iin()
        fct = factors(n)
        if len(fct)>=3:
            l = len(fct)
            a, b= fct[0], fct[1]
            c = 1
            if b==a:
                if l>3:
                    b*=fct[2]
                    for i in range(3, l):
                        c *= fct[i]
                else:
                    print(ans[1])
                    continue
            else:
                for i in range(2, l):
                    c *= fct[i]
            if a!=b and b!=c and c!=a:
                print(ans[0])
                print(a, b, c)
                continue
        print(ans[1])




            



            
        

        
main()
# threading.Thread(target=main).start()
