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 
#Author: ghoshashis545 -Ashis Ghosh
from __future__ import division, print_function
 
import os,sys
from io import BytesIO, IOBase
from math import ceil,sqrt
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
 
 
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def li(): return list(mi())
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[0] 
def sort2(l):return sorted(l, key=getKey)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
abc='abcdefghijklmnopqrstuvwxyz'
def powerMod(x,y,p):
    res = 1
    x %= p
    while y > 0:
        if y&1:
            res = (res*x)%p
        y = y>>1
        x = (x*x)%p
    return res
 
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x
    
def isPrime(n) : # Check Prime Number or not 
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) : 
        if (n % i == 0 or n % (i + 2) == 0) : 
            return False
        i = i + 6
    return True
def fact(n):
    f=1
    for i in range(1,n+1):
        f*=i
    return f
def f(n):
    a=[]
    a.append(1)
    for i in range(2,int(sqrt(n))+1):
        if(n%i==0):
            if(i==n//i):
                a.append(i)
            else:
                a.append(i)
                a.append(n//i)
    return a
''''''''''''''''''''''''''''''''''''''''''''''''''''''
t=ii()
while(t):
    t-=1
    n=ii()
    f=0
    for i in range(2,int(sqrt(n))+1):
        if(n%i==0):
            a=[]
            a.append(i)
            x=n//i
            for j in range(2,int(sqrt(x))+1):
                if(x%j==0 and j not in a):
                    a.append(j)
                    a.append(x//j)
                    if(len(set(a))==3):
                        print("YES")
                        print(*a)
                        f=1
                        break
            if(f==1):
                break
            a=[]
            a.append(n//i)
            x=i
            for j in range(2,int(sqrt(x))+1):
                if(x%j==0 and j not in a):
                    a.append(j)
                    a.append(x//j)
                    if(len(set(a))==3):
                        print("YES")
                        print(*a)
                        f=1
                        break
            if(f==1):
                break
    if(f==0):
        print("NO")
    
    
    
    
