Problem: Today the kindergarten has a new group of n kids who need to be seated at the dinner table. The chairs at the table are numbered from 1 to 4n. Two kids can't sit on the same chair. It is known that two kids who sit on chairs with numbers a and b (a ≠ b) will indulge if: 

  1. gcd(a, b) = 1 or, 
  2. a divides b or b divides a. 



gcd(a, b) — the maximum number x such that a is divisible by x and b is divisible by x.

For example, if n=3 and the kids sit on chairs with numbers 2, 3, 4, then they will indulge since 4 is divided by 2 and gcd(2, 3) = 1. If kids sit on chairs with numbers 4, 6, 10, then they will not indulge.

The teacher really doesn't want the mess at the table, so she wants to seat the kids so there are no 2 of the kid that can indulge. More formally, she wants no pair of chairs a and b that the kids occupy to fulfill the condition above.

Since the teacher is very busy with the entertainment of the kids, she asked you to solve this problem.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

Each test case consists of one line containing an integer n (1 ≤ n ≤ 100) — the number of kids.

Output

Output t lines, which contain n distinct integers from 1 to 4n — the numbers of chairs that the kids should occupy in the corresponding test case. If there are multiple answers, print any of them. You can print n numbers in any order.

Example

Input


3
2
3
4


Output


6 4
4 6 10
14 10 12 8

Solution (in PYTHON3):

Solution:
#import modules here
import math,sys,os
#from itertools import permutations, combinations
# from collections import defaultdict,deque,OrderedDict,Counter
# import bisect as bi
#import heapq
from io import BytesIO, IOBase
mod=10**9+7

# region fastio
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")

#input functions
def minp(): return map(int, sys.stdin.readline().rstrip().split())
def linp(): return list(map(int, sys.stdin.readline().rstrip().split()))
def inp(): return int(input())
def print_list(l):
    print(' '.join(map(str,l)))

# functions
# def BinarySearch(a,x): 
#     i=bi.bisect_left(a, x) 
#     if i != len(a) and a[i] == x: 
#         return i 
#     else: 
#         return -1
        
# def gcd(a,b):
#     return math.gcd(a,b)
    
# def is_prime(n):
#     """returns True if n is prime else False"""
#     if n < 5 or n & 1 == 0 or n % 3 == 0:
#         return 2 <= n <= 3
#     s = ((n - 1) & (1 - n)).bit_length() - 1
#     d = n >> s
#     for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
#         p = pow(a, d, n)
#         if p == 1 or p == n - 1 or a % n == 0:
#             continue
#         for _ in range(s):
#             p = (p * p) % n
#             if p == n - 1:
#                 break
#         else:
#             return False
#     return True
    
# def mulinverse(a):
#     return pow(a,mod-2,mod)

####################Let's Go Baby########################
for _ in range(inp()):
    n=inp()
    start=2*n
    for i in range(n):
        print(start,end=" ")
        start+=2
    print()