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 
# Python3 implementation of the approach 
from math import sqrt


# Function to find the required triplets
def findTriplets(x):
    # To store the factors
    fact = [];
    factors = set();

    # Find factors in sqrt(x) time
    for i in range(2, int(sqrt(x))):
        if (x % i == 0):
            fact.append(i);

            if (x / i != i):
                fact.append(x // i);

            factors.add(i);
            factors.add(x // i);

    found = False;
    k = len(fact);

    for i in range(k):

        # Choose a factor
        a = fact[i];

        for j in range(k):

            # Choose another factor
            b = fact[j];

            # These conditions need to be
            # met for a valid triplet
            if ((a != b) and (x % (a * b) == 0)
                    and (x / (a * b) != a)
                    and (x / (a * b) != b)
                    and (x / (a * b) != 1)):
                # Print the valid triplet
                print("YES")
                print(a, b, x // (a * b));
                found = True;
                break;

            # Triplet found
        if (found):
            break;

        # Triplet not found
    if (not found):
        print("NO");

    # Driver code


if __name__ == "__main__":
    t=int(input())
    while(t):
        t-=1
        n=int(input())
        findTriplets(n)


