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 math import *
input=sys.stdin.readline
from math import *
t=int(input())
while t>0:
    t-=1
    n=int(input())
    d={}
    for i in range(2,int(ceil(sqrt(n)))+1):
        if n%i==0:
            d[i]=0
            while n%i==0:
                n//=i
                d[i]+=1
    if n>2:
        d[n]=1

      
    if len(d)>=3:
        a=[]
        for i in d:
            a.append(i**d[i])
        a.sort(reverse=True)    
        p=a[0]
        q=a[1]
        s=1
        for k in range(2,len(a)):
            s*=a[k]
        print("YES")
        print(p,q,s)
        continue
    elif len(d)==2:
        a=[]
        s=0
        for i in d:
            s+=d[i]
        if s<=3:
            print("NO")
            continue
        s=1    
        for i in d:
            a.append(i)
            s*=i**(d[i]-1)
        a.sort()
        p=a[0]
        q=a[1]
      
        print("YES")
        print(p,q,s)
            
    else:
        c=0
        a=[]
        for i in d:
            if 6>d[i]>=3:
                a.append(i)
                d[i]-=1
                a.append(i**(d[i]))
            elif d[i]>=6:    
                a.append(i)
                a.append(i**2)
                a.append(i**(d[i]-3))
            else:
                a.append(i**d[i])
        
        if len(a)<3:
            print("NO")
            continue
        a.sort(reverse=True)    
        p=a[0]
        q=a[1]
        s=1
        for k in range(2,len(a)):
            s*=a[k]
        print("YES")
        print(p,q,s)
        continue