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__ = 'Devesh Bajpai'

'''
https://codeforces.com/problemset/problem/1294/C

Solution: We first find the factor for n using a standard sqrt time complexity stub. If we could 
find one, that becomes a. And then we find factor for n/a which is not equal to a. If we could
find oen, that becomes b. By then, the value of n is made (n/a)/b or n/ab. Hence if that value of
n is not equal to a,b or 1 then it is c. We add it to the distinct factor's set and return the result.
At any step if we fail, the answer returned is NO. 
'''


def solve(n):

    distinct_factors = set()

    done, a = find_factors(n, distinct_factors)
    if not done:
        return "NO"
    else:
        n /= a

    done, b = find_factors(n, distinct_factors)
    if not done:
        return "NO"
    else:
        n /= b

    if n in distinct_factors or n == 1:
        return "NO"
    else:
        distinct_factors.add(n)

    return "YES" + "\n" + " ".join(str(_) for _ in distinct_factors)


'''
Helper to find distinct factor for the current value of n
Time Complexity: O(sqrt(n))
'''


def find_factors(n, distinct_factors):

    factor = 2
    f = 2
    done = False
    while f * f <= n and not done:

        if n % f == 0 and f not in distinct_factors:

            distinct_factors.add(f)
            factor = f
            done = True

        f += 1

    return done, factor


if __name__ == "__main__":

    t = int(raw_input())

    results = list()
    for _ in xrange(0, t):
        n = int(raw_input())
        results.append(solve(n))

    for result in results:
        print result
