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 sys
from collections import defaultdict
import math

def primeFactors(n):
    # Print the number of two's that divide n
    ans=[]
    while n % 2 == 0:
        ans.append(2)
        n = n / 2

    # n must be odd at this point
    # so a skip of 2 ( i = i + 2) can be used
    for i in range(3, int(math.sqrt(n)) + 1, 2):

        # while i divides n , print i ad divide n
        while n % i == 0:
            ans.append(i)
            n = n / i

    # Condition if n is a prime
    # number greater than 2
    if n > 2:
        ans.append(int(n))

    return ans

t = int(input())
for _ in range(t):
    n=int(input())
    #n,k=map(int,input().split())
    #b=list(map(int,input().split()))
    req=primeFactors(n)
    d=defaultdict(lambda:0)
    for j in req:
        d[j]+=1

    l=sorted(list(set(req)))
    if len(l)>=3:
        print("YES")
        print(l[0], l[1], n//(l[0]*l[1]))

    elif len(l)==2:
        ans=[l[0],l[1],n//(l[0]*l[1])]
        if len(set(ans))==3 and min(ans)!=1:
            print("YES")
            print(l[0], l[1], n // (l[0] * l[1]))

        else:
            print("NO")

    else:
        if d[l[0]]<6:
            print("NO")

        else:
            print('YES')
            print(l[0], l[0]**2, n//(l[0]**3))














