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 primeFactors(n):
    mm=n+1
    d=dict()
    count=0
    while n % 2 == 0: 
        count+=1
        n = n // 2
    if count!=0:
        d[2]=count
    # print("Intermediate n",n)
    for i in range(3,int(math.sqrt(n))+1,2):
        icount=0
        while n % i== 0: 
            # print i, 
            n = n // i
            icount+=1
            # print("Andar")
        # print("Intermediate icount",icount)
        if icount!=0:
            d[i]=icount
            # print("Assigned")
    if n > 2: 
        # print n
        d[n]=1
    n=mm-1
    li=[]
    for i in d:
        # print(i,d[i])
        li.append(i)
    # print(len(d))
    if len(d)>=3:
        print("YES")
        num1=li[0]
        num2=li[1]
        num3=n//num1
        num3=num3//num2
        print(num1,num2,num3)
    elif len(d)==2:
        anss=0
        for i in d:
            anss+=d[i]
        if anss>=4:
            print("YES")
            fi=li[0]
            se=li[1]
            print(fi,se,(n//fi)//se)
        else:
            print("NO")
    else:
        anss=0
        for i in d:
            anss+=d[i]
        if anss>=6:
            print("YES")
            num1=li[0]
            num2=num1**2
            num3=num1**3
            print(num1,num2,(n//num1)//num2)
        else:
            print("NO")

t=int(input())
for i in range(0,t):
    n=int(input())
    primeFactors(n)