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 math
from collections import Counter


def primeFactors(n):
    # Print the number of two's that divide n
    l = []
    while n % 2 == 0:
        l.append(2)
        n = n / 2

    # n must be odd at this point
    # so a skip of 2 ( i = i + 2) can be used
    for i in range(3, int(math.sqrt(n)) + 1, 2):

        # while i divides n , print i ad divide n
        while n % i == 0:
            l.append(int(i))
            n = n / i

    # Condition if n is a prime
    # number greater than 2
    if n > 2:
        l.append(int(n))
    return l, Counter(l)


if __name__ == '__main__':
    for _ in range(int(input().strip())):
        n = int(input().strip())
        ll, pf = primeFactors(n)
        ll.sort()
        if len(pf) >= 3:
            print("YES")
            a = ll[0]
            b = ll[-1]
            c = 1
            for i in range(1, len(ll) - 1):
                c *= ll[i]
            print(int(a), int(b), int(c))
        elif len(pf) == 2:
            if len(ll) >= 4:
                print("YES")
                a = ll[0]
                b = ll[-1]
                c = 1
                for i in range(1, len(ll) - 1):
                    c *= ll[i]
                print(int(a), int(b), int(c))
            else:
                print("NO")
        else:
            if len(ll) >= 6:
                print("YES")
                tot = ll[0] ** len(ll)
                print(ll[0], ll[0] ** 2, (tot // (ll[0] * (ll[0] * ll[0]))))
            else:
                print("NO")
