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
from collections import Counter
def primeFactors(n):
    a = []
    c = 0
    while n % 2 == 0:
        a.append(2)
        n = n // 2
        c+=1
    # 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:
            a.append(i)
            n = n // i
            c+=1
            # Condition if n is a prime
    # number greater than 2
    if n > 2:
        a.append(n)
        c+=1
    return Counter(a),c

cases = int(input())
for t in range(cases):
    n = int(input())
    d,c = primeFactors(n)
    v = len(d)
    if v>2:
        v = list(d.keys())
        print("YES")
        print(v[0],v[1],n//(v[0]*v[1]))
    elif v==2:
        if c>3:
            v = list(d.keys())
            print("YES")
            print(v[0],v[1],n//(v[0]*v[1]))
        else:
            print("NO")
    else:
        if c>5:
            v = list(d.keys())[0]
            print("YES")
            print(v,v**2,n//(v**3))
        else:
            print("NO")
