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,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right 
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()

def prime_factorize(n):
    if n == 1:
        return [1]
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a
for _ in range(inp()):
    n = inp()
    pp = prime_factorize(n)
    c = Counter(pp)
    bad = False
    if (len(c.keys()) == 1 and len(pp)<6) or (len(c.keys())==2 and len(pp)<4):
        print('NO')
    else:
        keys = list(c.keys())
        if len(c.keys()) > 1:
            a = keys[0]; c[keys[0]] -= 1
            b = keys[1]; c[keys[1]] -= 1
            C = 1
            for key in keys:
                if c[key] == 0: continue
                C *= key**c[key]
        else:
            a = keys[0]*1; b = keys[0]**2
            C = keys[0]**(c[keys[0]]-3) 
        print('YES')
        print(a,b,C)