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


def divisors(n):
    divs = list()
    for i in range(2, int(math.sqrt(n)) + 1):
        if n % i == 0:
            divs.extend([i, n // i])
    return list(sorted(set(divs)))


def valid(first, second, third, result):
    return third > 1 and third != first and third != second and first * second * third == result


n = int(input())

for _ in range(n):
    value = int(input())
    divs = divisors(value)
    temp = value

    if len(divs) == 0:
        print("NO")
        continue

    if len(divs) == 1:

        first = divs[0]
        second = first * first
        third = value // (first * second)

        if valid(first, second, third, value):
            print("YES")
            print(first, second, third)
        else:
            print("NO")

        continue

    found = False

    for i in range(len(divs)):
        for j in range(i + 1, len(divs)):
            first, second = divs[i], divs[j]

            if valid(first, second, value // (first * second), value):
                print("YES")
                print(first, second, value // (first * second))
                found = True
            elif valid(first, first * first, value // (first ** 3), value):
                print("YES")
                print(first, first * first, value // (first ** 3))
                found = True

            if found:
                break

        if found:
            break

    if not found:
        print("NO")
