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 primeFactors(n):
    vals = []
    # Print the number of two's that divide n
    two = False
    while n % 2 == 0:
        vals.append(int(n))
        if not two:
            vals.append(2)
            two = True
        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:
            vals.append(int(n))
            vals.append(i)
            n = n / i

            # Condition if n is a prime
    # number greater than 2
    if n > 2:
        vals.append(int(n))
    if len(vals) != 0:
        vals.pop(0)
    vals.sort()
    i = 1
    while i < len(vals):
        if vals[i-1] == vals[i]:
            del vals[i]
        else:
            i+= 1
    return vals

def find3(number):
    factors = primeFactors(number)
    if len(factors) == 0:
        return []
    finalFactors = [factors[0]]
    newNumber = int(number/factors[0])
    factors = primeFactors(newNumber)
    if len(factors) == 0:
        return finalFactors
    pos = 0
    if factors[pos] == finalFactors[0]:
        newFound = False
        while not newFound:
            if factors[pos] != finalFactors[0]:
                newFound = True
            else:
                pos += 1
                if pos >= len(factors):
                    return finalFactors

    if len(factors) == 0:
        finalFactors.append(newNumber)
        return finalFactors
    finalFactors.append(factors[pos])

    newNumber = int(newNumber/factors[pos])
    for elem in finalFactors:
        if newNumber == elem:
            return finalFactors
    finalFactors.append(newNumber)
    return finalFactors

def solve(number):
    ans = find3(number)
    for i in range(len(ans)):
        ans[i] = str(ans[i])
    if len(ans) == 3:
        print("YES")
        p = " ".join(ans)
        print(p)
    else:
        print("NO")

input()
while True:
    try:
        val = int(input())
        solve(val)
    except EOFError:
        break
