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 collections
import itertools
import sys
from collections import defaultdict, Counter
from math import sqrt, ceil

input = sys.stdin.readline


############ ---- Input Functions ---- ############
def inp():
    return (int(input()))


def inlt():
    return (list(map(int, input().split())))


def insr():
    s = input()
    return (list(s[:len(s) - 1]))


def invr():
    return (map(int, input().split()))


def ds(d, s):
    for i in range(s, d + 1):
        if i*i>d:
            break
        if d % i == 0:
            return i


def main():
    n = inp()
    for _ in range(n):
        d = inp()
        d1 = ds(d, 2)


        if d1 is not None:
            d2 = ds(d // d1, d1 + 1)
            if d2 is not None:
                d3 = d // d1 // d2
                if d3 != 1 and d3 != d1 and d3 != d2:
                    print("YES")
                    print(" ".join([str(v) for v in [d1, d2, d3]]))
                else:
                    print("NO")
            else:
                print("NO")
        else:
            print("NO")


if __name__ == "__main__":
    # sys.setrecursionlimit(10 ** 6)
    # threading.stack_size(10 ** 8)
    # t = threading.Thread(target=main)
    # t.start()
    # t.join()
    main()
