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

def factor(n,a):

    l=[]
    r=n
    k=int(math.sqrt(n))+1

    d=1
    
    while n%(2**d)==0:
        if 2**d!=a and r//(2**d)!=a and r//(2**d)!=2**d and 2**d!=r:
            l.append(2**d)
            l.append(r//2**d)
            
            n//=2**d
            d+=1
            break
        
        else:
            d+=1


    for i in range(3,k,2):
        d=1
        while n%(i**d)==0:
            if i**d!=a and r//(i**d)!=a and r//(i**d)!=i**d and i**d!=n:
                l.append(i**d)
                
                l.append(r//i**d)
                n//=i**d
                d+=1
                break
            else:
                d+=1

    return l

def factors(n):

    l=[]
    r=n
    k=int(math.sqrt(n))+1

    d=1
    while n%2==0:
        l.append(2**d)
        d+=1

        n//=2

    for i in range(3,k,2):
        d=1
        while n%i==0:
            l.append(i**d)
            d+=1
            
            n//=i

    if n==1:
        return l
    else:
        l.append(n)
        return l
    
for _ in range(int(input())):

    n=int(input())

    l=factors(n)

    if len(l)<3:
        print("NO")

    else:
        t=x=y=z=0
        for i in range(len(l)):
            a=n//l[i]
            
            f=factor(a,l[i])
            if len(f)>1:
                x=f[0]
                y=f[1]
                z=l[i]
                t=1
                break

        if t==1:
            print("YES")
            print(z,x,y)

        else:
            print("NO")
