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:
# ------------------- fast io --------------------
import os
import sys
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")
 
# ------------------- fast io --------------------

import math
import bisect
primes=[2]
for j in range(3,10**5):
    indy=min(bisect.bisect_left(primes,math.ceil(math.sqrt(j))),len(primes)-1)
    broke=False
    for s in range(indy+1):
        if j%primes[s]==0:
            broke=True
            break
    if broke==False:
        primes.append(j)
minny=len(primes)
testcases=int(input())

for j in range(testcases):
    #ok we will find its prime divisors
    n=int(input())
    indy=min(bisect.bisect_left(primes,math.ceil(math.sqrt(n))),minny-1)
    facs=[]
    for s in range(indy+1):
        if n%primes[s]==0:
            facs.append(primes[s])
    exi=True
    if len(facs)>=3:
        a,b,c=facs[0],facs[1],n//(facs[0]*facs[1])
    elif len(facs)==2:
        a,b,c=facs[0],facs[1],n//(facs[0]*facs[1])
        if c==0 or c==1 or c==a or c==b:
            exi=False
    elif len(facs)==1:
        a,b,c=facs[0],facs[0]**2,n//(facs[0]**3)
        if c==0 or c==1 or c==a or c==b:
            exi=False
    elif len(facs)==0:
        a,b,c=0,0,0
        exi=False
    if a*b*c==n and exi==True:
        print("YES")
        print(str(a)+" "+str(b)+" "+str(c))
    else:
        print("NO")