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
def fastio():
    from io import StringIO
    from atexit import register
    global input
    sys.stdin = StringIO(sys.stdin.read())
    input = lambda : sys.stdin.readline().rstrip('\r\n')
    sys.stdout = StringIO()
    register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()

def debug(*var, sep = ' ', end = '\n'):
    print(*var, file=sys.stderr, end = end, sep = sep)

INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br

def ok(n):
    # for i in range()
    pass

def fac(n):
    res = []
    for i in range(2, int(n ** 0.5 + 1)):
        if n % i:
            continue
        res.append(i)
        if i ** 2 != n:
            res.append(n // i)
    return sorted(res)

t, = I()
while t:
    t -= 1
    n, = I()
    ans = 'YES'
    f = fac(n)
    if len(f) < 3:
        print('NO')
        continue
    m = len(f)
    for i in range(m):
        for j in range(i + 1, m):
            k = n // (f[i] * f[j])
            if k in f and k != f[i] and k != f[j]:
                print('YES')
                print(f[i], f[j], k)
                break
        else:
            continue
        break
    else:
        print('NO')