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,sys
from collections import Counter, defaultdict, deque
from sys import stdin, stdout
input = stdin.readline
lili=lambda:list(map(int,sys.stdin.readlines()))
li = lambda:list(map(int,input().split()))
#for deque append(),pop(),appendleft(),popleft(),count()
I=lambda:int(input())
S=lambda:input().strip()
mod = 1000000007

# A Better (than Naive) Solution to find all divisiors 
import math 

# method to print the divisors 
def printDivisors(n) : 

    a=[]
    i = 1
    while i <= math.sqrt(n): 

        if (n % i == 0) : 

        # If divisors are equal, print only one 
            if (n / i == i) : 
                a.append(i)
            else : 
            # Otherwise print both 
                a.append(i)
                a.append(n//i)
        i = i + 1
    return (a)
for i in range(I()):
    n=I()
    a=printDivisors(n)
    d=Counter(a)
    for i in a:
        f=0
        if(i==1 or i==n):
            continue
        p=n//i
        #print(p)
        j=1
        while(j<=math.sqrt(p)):
            if(p%j==0 and j!=1 and j in d and (p//j)  in d and j!=i and i!=(p//j) and j!=(p//j) and j!=p):
                print("YES")
                print(i,j,p//j)
                f=1
                break
            j+=1
        if(f==1):
            break

    if(f==0):
        print("NO")

