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 sys
from math import ceil, sqrt

t = int(sys.stdin.readline())

def factor(n):
    pf = {}
    maxp = min(n - 1, ceil(sqrt(n)))
    nf = n
    for p in range(2, maxp + 1):
        if nf == 1:
            break
        while nf % p == 0:
            nf //= p
            if p in pf:
                pf[p] += 1
            else:
                pf[p] = 1
    if nf > 1:
        if nf in pf:
            pf[nf] += 1
        else:
            pf[nf] = 1
    
    pf = list(pf.items())
    # ~ print(n, pf)
    if len(pf) == 1:
        if pf[0][1] >= 6:
            print("YES")
            frac = n // pf[0][0]**3
            print(pf[0][0], pf[0][0]**2, frac)
        else:
            print("NO")
    elif len(pf) == 2:
        if pf[0][1] >= 2 and pf[1][1] >= 2:
            print("YES")
            frac = n // (pf[0][0] * pf[1][0])
            print(pf[0][0], pf[1][0], frac)
        elif pf[0][1] == 1 and pf[1][1] >= 3 or pf[0][1] >= 3 and pf[1][1] == 1:
            print("YES")
            frac = n // (pf[0][0] * pf[1][0])
            print(pf[0][0], pf[1][0], frac)
        else:
            print("NO")
    else:
        print("YES")
        frac = n // (pf[0][0] * pf[1][0]) 
        print(pf[0][0], pf[1][0], frac)


for i in range(t):
    n = int(sys.stdin.readline())
    factor(n)
