Problem: 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 

Solution (in PYTHON3):

Solution:
def factorization(num):
    pn = []
    while num % 2 == 0:
        pn.append(2)
        num //= 2
    while num % 3 == 0:
        pn.append(3)
        num //= 3
    fg = False
    for i in range(5, int(num ** 0.5) + 1, 6):
        if fg:
            if i > int(num ** 0.5):
                break
            fg = False
        while num % i == 0:
            pn.append(i)
            num //= i
            fg = True
        i += 2
        if fg:
            if i > int(num ** 0.5):
                break
            fg = False
        while num % i == 0:
            pn.append(i)
            num //= i
            fg = True
    if num > 1:
        pn.append(num)
    return pn


def calculation():
    if len(ns) < 3:
        return 0, 0, 0
    n1 = ns[0]
    n2 = ns[1]
    n3 = 0
    for i in range(2, len(ns)):
        if n2 == n1:
            n2 *= ns[i]
            break
        i -= 1
        break
    if len(ns) > i + 1:
        n3 = ns[i + 1]
    else:
        return 0, 0, 0
    for i in range(i + 2, len(ns)):
        n3 *= ns[i]
    if n1 != n2 and n2 != n3 and n1 != n3:
        return n1, n2, n3
    return 0, 0, 0


t = int(input())
for _ in range(t):
    n = int(input())
    if n > 23:
        ns = factorization(n)
        n1, n2, n3 = calculation()
        if n1 > 0:
            print('YES')
            print(n1, n2, n3)
        else:
            print('NO')
    else:
        print('NO')