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 
"""
Author : Aman Thakur
mantra: chor kya hi kar sakte hai!!
"""
import math


class Solution:
    def __init__(self):
        self.c = 0
        self.arr = []

    # def solution(self):
        # n = int(input())
        #
        # for i in range(2, int(math.sqrt(n)) + 1):
        #     if n % i == 0:
        #         self.arr.append(i)
        #         n //= i
        #         break
        #
        # for i in range(self.arr[0] + 1, int(math.sqrt(n)) + 1):
        #     if n % i == 0 and n // i not in self.arr and n // i > 1:
        #         self.arr.append(i)
        #         if i != n // i:
        #             self.arr.append(n // i)
        #         break
        #
        # if len(self.arr) == 3:
        #     print('YES')
        #     for i in self.arr:
        #         print(i, end=' ')
        #     print()
        # else:
        #     print('NO')
    def solution(self):
        n = int(input())

        for i in range(2, n):
            if n % i == 0:
                self.arr.append(i)
                n //= i
            if len(self.arr) == 2 or n < i*i:
                break

        if len(self.arr) == 2 and n != self.arr[0] and n != self.arr[1] and n > 1:
            print('YES')
            print(self.arr[0], self.arr[1], n)
        else:
            print('NO')


if __name__ == '__main__':
    for _ in range(int(input())):
        ob = Solution()
        ob.solution()
