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 
from math import floor, sqrt, ceil
def findPrimeFac(num):
    ans = []
    lim = ceil(sqrt(num))
    while num % 2 == 0:
        ans.append(2)
        num //= 2
    for p in range(3, lim, 2):
        while num % p == 0:
            ans.append(p)
            num //= p
    if num != 1:
        ans.append(num)
    return ans

t = int(input())
for _ in range(t):
    n = int(input())
    primeFacs = findPrimeFac(n)
    #print(primeFacs)
    freq = {}
    for p in primeFacs:
        if p not in freq:
            freq[p] = 0
        freq[p] += 1
    numDist = 0
    ans = []
    l = list(freq.keys())
    if len(l) >= 2:
        a = l[0]
        b = l[1]
        c = n // (a * b)
        if c != a and c != b and c != 1:
            print("YES")
            assert n == a*b*c
            print(a, b, c)
        else:
            print("NO")
    elif len(l) == 0 or freq[l[0]] < 6:
        print("NO")
    else:
        print("YES")
        p = l[0]
        print(p, p*p, n//(p ** 3))
    

            

