Problem: 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 

Solution (in PYTHON3):

Solution:
import math
def primefact(n):
    c = 0
    l=[]
    while n % 2 == 0:
        c = c + 1
        l.append(2)
        n = n / 2

    for i in range(3, int(math.sqrt(n)) + 1, 2):
        while n % i == 0:
            c = c + 1
            l.append(i)
            n = n / i

    if n > 2:
        l.append(int(n))
        c = c + 1

    return l
t=int(input())
for i in range(t):
    temp=[]
    n=int(input())
    temp=primefact(n)
    dic={}
    for j in temp:
        if j in dic:
            dic[j]=dic[j]+1
        else:
            dic[j]=1

    if len(dic)==1:
        for k in dic:
            if dic[k]>=6:
                print("YES")
                print(k*1,k*2,int(math.pow(k,(dic[k]-3))))
            else:
                print("NO")
    elif len(dic)==2:
       k=list(dic.keys())
       if dic[k[0]]>=2 and dic[k[1]]>=2:
           print("YES")
           print(k[0],int(math.pow(k[0],(dic[k[0]]-1))*math.pow(k[1],(dic[k[1]]-1))),k[1])
       elif dic[k[0]]>=3:
           print("YES")
           print(k[0],int(math.pow(k[0],(dic[k[0]]-1))),k[1])
       elif dic[k[1]] >= 3:
           print("YES")
           print(k[0],int(math.pow(k[1], (dic[k[1]] - 1))), k[1])
       else:
           print("NO")
    elif len(dic)>=3:
        print("YES")
        k = list(dic.keys())
        print(int(math.pow(k[0],(dic[k[0]]))),end=" ")
        print(int(math.pow(k[1],(dic[k[1]]))),end=" ")
        p = 1
        for j in range(2, len(k)):
            p = p * math.pow(k[j], (dic[k[j]]))
        print(int(p))
    else:
        print("NO")