Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.

Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer x as nearly prime if it can be represented as p ⋅ q, where 1 < p < q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2 ⋅ 3 = 6 and 2 ⋅ 5 = 10), but integers 1, 3, 4, 16, 17 or 44 are not.

Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positive integers where at least 3 of them should be nearly prime.

Uncle Bogdan easily solved the task and joined the crew. Can you do the same?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

Next t lines contain test cases — one per line. The first and only line of each test case contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number Flint guessed.

Output

For each test case print: 

  * YES and 4 different positive integers such that at least 3 of them are nearly prime and their sum is equal to n (if there are multiple answers print any of them); 
  * NO if there is no way to represent n as the sum of 4 different positive integers where at least 3 of them are nearly prime. 

You can print each character of YES or NO in any case.

Example

Input


7
7
23
31
36
44
100
258


Output


NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6

Note

In the first and second test cases, it can be proven that there are no four different positive integers such that at least three of them are nearly prime.

In the third test case, n=31=2 ⋅ 7 + 2 ⋅ 5 + 2 ⋅ 3 + 1: integers 14, 10, 6 are nearly prime.

In the fourth test case, n=36=5 + 2 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 5: integers 6, 10, 15 are nearly prime.

In the fifth test case, n=44=2 ⋅ 3 + 7 + 2 ⋅ 5 + 3 ⋅ 7: integers 6, 10, 21 are nearly prime.

In the sixth test case, n=100=2 + 2 ⋅ 5 + 3 ⋅ 11 + 5 ⋅ 11: integers 10, 33, 55 are nearly prime.

In the seventh test case, n=258=2 ⋅ 5 + 3 ⋅ 7 + 13 ⋅ 17 + 2 ⋅ 3: integers 10, 21, 221, 6 are nearly prime.
"""
    Satwik_Tiwari ;) .
    30th july , 2020  - Thursday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase

from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================
#fast I/O region
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")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
    ans = 1
    while(b>0):
        if(b%2==1):
            ans*=a
        a*=a
        b//=2
    return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) :
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))

def solve():
    n = int(inp())
    if(n>30):
        if(n-30==6):
            print('YES')
            print(6,10,15,n-31)
        elif(n-30==10):
            print('YES')
            print(6,14,15,n-35)
        elif(n-30==14):
            print('YES')
            print(6,10,15,n-31)
        else:
            print('YES')
            print(6,10,14,n-30)
    else:
        print('NO')




# testcase(1)
testcase(int(inp()))
