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
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict
from bisect import bisect_left as bl, bisect_right as br
from collections import Counter
from collections import deque
 
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()

t = ip()
for _ in range(t):
    n = ip()
    ch = []
    for i in range(2,int(n**0.5)+1):
        if n%i == 0:
            ch.append(i)
            ch.append(n//i)
            break
    if len(ch) == 2:
        a = ch[0]
        b = ch[1]
        ch = []
        for i in range(2,int(a**0.5)+1):
            if a%i == 0:
                if i != b and a//i != b and i != a//i:
                    ch.append(i)
                    ch.append(a//i)
                    break
        if len(ch) == 2:
            print("YES")
            print(b,ch[0],ch[1])
        else:
            ch = []
            for i in range(2,int(b**0.5)+1):
                if b%i == 0:
                    if i != a and b//i != a and i!= b//i:
                        ch.append(i)
                        ch.append(b//i)
                        break
            if len(ch) == 2:
                print("YES")
                print(a,ch[0],ch[1])
            else:
                print("NO")
    else:
        print("NO")
        
        
            
            
            
        
