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 
'''

* Author : Ayushman Chahar #
* About  : II Year, IT Undergrad #
* Insti  : VIT, Vellore #

'''

import os
import sys
import math
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")


def pf(n):
    ans = []
    while(n % 2 == 0):
        ans.append(2)
        n //= 2
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        while(n % i == 0):
            ans.append(i)
            n //= i
    if(n > 2):
        ans.append(n)
    return ans

def main():
    for _ in range(int(input())):
        n = int(input())
        res = pf(n)
        # print(res)
        if(len(res) <= 2):
            print("NO")
        else:
            if(len(res) == 3):
                if(len(set(res)) == 3):
                    print("YES")
                    print(*res)
                else:
                    print("NO")
            else:
                a1 = res[0]
                a2 = 1
                idx = -69
                for i in range(1, len(res)):
                    if a2 != a1 and a2 != 1:
                        idx = i
                        break
                    else:
                        a2 *= res[i]
                a3 = 1
                for i in range(idx, len(res)):
                    a3 *= res[i]
                if(a3 != a2 and a3 != a1):
                    print("YES")
                    print(a1, a2, a3)
                else:
                    print("NO")
                    
if __name__ == "__main__":
    main()
