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 bisect import bisect_left, bisect_right
from sys import stdin, stdout
input = lambda: stdin.readline().strip()
print = lambda s: stdout.write(s)

primes = []
def SieveOfEratosthenes(n):
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    for p in range(2, n):
        if prime[p]:
            primes.append(p)
SieveOfEratosthenes(10**6)

t = int(input())
for _ in range(t):
    n = int(input())
    ls = []
    for i in primes:
        if n%i==0:
            ls.append(i)
        if len(ls)==2:
            break
    if len(ls)==0:
        print('NO\n')
    elif len(ls)>=2:
        if n//(ls[0]*ls[1]) in ls or n//(ls[0]*ls[1])==1:
            print('NO\n')
        else:
            print('YES\n')
            print(str(ls[0])+' '+str(ls[1])+' '+str(n//(ls[0]*ls[1]))+'\n')
    elif n==ls[0] or n==ls[0]**2 or n==ls[0]**3 or n==ls[0]**4 or n==ls[0]**5:
        print('NO\n')
    else:
        copy = n
        cnt = 0
        while copy%ls[0]==0:
            copy//=ls[0]
            cnt+=1
        if cnt>=3:
            print('YES\n')
            print(str(ls[0])+' '+str(ls[0]**2)+' '+str(n//(ls[0]**3))+'\n')
        else:
            print('NO\n')
