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 primeFactors(n): 
    li =[]  
    while n % 2 == 0:
        li.append(2)
        #print (2) 
        n = n//2

    for i in range(3,int(math.sqrt(n))+1,2): 

        while n % i== 0: 
            ##print (i)
            li.append(i)
            n = n//i 
              
    if n > 2:
        li.append(n)
        #print (n)
    return li

for _ in range(int(input())):
    n = int(input())
    li = primeFactors(n)
    d = dict()
    for i in li:
        if i not in d:
            d[i] = 1
        else:
            d[i] += 1
    #print(d)
    if(len(d) == 1):
        x = li[0]
        if(d[x]>=6):
            print("YES")
            print(x,x*x,pow(x,d[x]-3))
        else:
            print("NO")
    if(len(d)>=3):
        x = 1
        lee = []
        for i  in d:
            lee.append((i,d[i]))
        a = pow(lee[0][0],lee[0][1])
        b = pow(lee[1][0],lee[1][1])
        c = 1
        for i in range(2,len(d)):
            c = c*pow(lee[i][0],lee[i][1])
        print("YES")
        print(a,b,c)
    if(len(d) ==2):
        lee = []
        for i  in d:
            lee.append((i,d[i]))
        if(lee[0][1]>=3):
            print("YES")
            x= lee[0][1]
            print(lee[0][0], pow(lee[0][0], lee[0][1]-1), pow(lee[1][0],lee[1][1]))
        elif(lee[1][1]>=3):
            print("YES")
            x= lee[1][1]
            print(pow(lee[0][0],lee[0][1]), lee[1][0], pow(lee[1][0], lee[1][1]-1))
        elif(lee[0][1] == 2 and lee[1][1]==2):
            print("YES")
            print(lee[0][0],lee[0][0]*lee[1][0], lee[1][0])
        else:
            print("NO")
            
        
            
