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 
"""
-----------------------------Pseudo---------------------------------
"""
import copy
import sys
from collections import defaultdict, Counter
#sys.setrecursionlimit(20000)
#PI = 3.1415926535897932384626433832795
def input(): return sys.stdin.readline()
def mapi(): return map(int,input().split())
def maps(): return map(str,input().split())
#
def print(arg, *argv, end=None):
    sys.stdout.write(str(arg))
    for i in argv: sys.stdout.write(" "+str(i))
    sys.stdout.write(end) if end else sys.stdout.write("\n")
#
def GCD(x, y): return GCD(y,x%y) if y else x
#
def modPow(x, y, p):
    res,x = 1,x%p
    while(y>0):
        if(y&1)==1: res=(res*x)%p
        y,x = y>>1,(x*x)%p
    return res
def modInv(s, mod): return modPow(s,mod-2,mod)
#---------------------------------------------------------------#

from math import *

def solve():
    t = 1
    t = int(input())
    while(t):
        t-=1
        n =  int(input())
        a = 0
        b = 0
        flag = 0
        for i in range(int(pow(n,1/3))+1, 1,-1):
            if n%i == 0:
                a = i
                #print(a)
                tmp = n//a
                for j in range(int(sqrt(tmp))+1, 1,-1):
                    if (tmp)%j==0 and a!=j:
                        b = j
                        c = tmp//b
                        if a!=c and b!=c:
                            if c>=2:
                                flag = 1
                            break
        if flag:
            print("YES")
            print(a,b,c)
        else:
            print("NO")







#---------------------------------------------------------------#

if __name__ == '__main__':
    solve()

