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 collections import defaultdict
import math
dic=defaultdict(int)
t=int(sys.stdin.readline())
for _ in range(t):
    dic=defaultdict(int)
    m=int(sys.stdin.readline())
    d=m
    l=[]
    while m%2==0:
        m=m//2
        dic[2]+=1
    #print(m,'mm')
    for i in range(3,int(math.sqrt(d))+1,2):
        #print(i,m)
        while m%i==0:
            #print('yes',i)
            dic[i]+=1
            m=m//i
    if m>2:
        dic[m]+=1
    #print(dic,'dic')
    if len(dic)>=3:
        print("YES")
        l=[]
        for i in dic:
            l.append(i)
            if len(l)==2:
                break
        l.append(d//(l[0]*l[1]))
        print(*l)
    elif len(dic)==2:
        a=[]
        for i in dic:
            a.append(i)
        z=True
        if dic[a[0]]>2 and z:
            l=[a[0],a[0]**(dic[a[0]]-1)]
            l.append(d//(l[0]*l[1]))
            print("YES")
            print(*l)
            z=False
        elif dic[a[1]]>2 and z:
            l=[a[1],a[1]**(dic[a[1]]-1)]
            l.append(d//(l[0]*l[1]))
            print("YES")
            print(*l)
            z=False
        elif dic[a[1]]==dic[a[0]]==2 and z:
            l=[a[0],a[1],a[0]*a[1]]
            print("YES")
            print(*l)
            z=False
        else:
            print("NO")
    elif len(dic)==1:
        for i in dic:
            if dic[i]>5:
                print("YES")
                l=[i,i**2,d//(i**3)]
                print(*l)
            else:
                print("NO")
    else:
        print("NO")
    
