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 math
import sys
from collections import OrderedDict
##### graph implementation with adjacancy list#####
class Graph:
    def __init__(self,Nodes,is_directed=False):
        self.nodes=Nodes
        self.adj_list={}
        self.is_directed=is_directed
 
        for node in self.nodes:
            self.adj_list[node]=[]
            
    def add_edge(self,u,v):
        self.adj_list[u].append(v)
        if self.is_directed==False:
            self.adj_list[v].append(u)
 
    def print_graph(self):
        for node in self.nodes:
            print(node,"->",self.adj_list[node])
 
    def degree_node(self,node):
        return len(self.adj_list[node])
 
    def dfsUTIL(self,v,visited,parents=[]):
        #print(v,end=" ")
        visited[v]=True
        for i in self.adj_list[v]:
            if visited[i]==False:
                self.dfsUTIL(i,visited,parents)
                parents.append(i)
    
    def dfs(self,v):
        visited=[False]*(max(self.adj_list)+1)
        parents=[v]
        self.dfsUTIL(v,visited,parents)
        return parents
######factorial on n######
def factorial(n):
    if n==0 or n==1:
        return 1
    ans=1
    for i in range(2,n+1):
        ans*=i
    return ans
#####sorting a dictionary by the values#####
def dict_sort(ans):
    ans=sorted(ans.items(),reverse=True,key=lambda kv:(kv[1]))
    
##### naive method for testing prime or not O(n^.5)#####
def is_prime(n):
    if n==1:
        return 0
    for i in range(2,int(math.sqrt(n))+1):
        if n%i==0:
            return False
    return True

#####swap function#####
def swap(a,b):
    temp=a
    a=b
    b=temp
    return a,b

#####Primes till Nth O(n)#####
def seive_primes(n):
    flag=[0]*(n+2)
    flag[1]=flag[0]=1
    i=2
    while i*i<=n+1:
        if flag[i]==0:
            j=i*i
            while j<=n+1:
                flag[j]=1
                j+=i
        i+=1
    return flag

#####all the prime factors of a number#####
def factors(n):
    d={}
    while(n%2==0):
        if 2 in d:
            d[2]+=1
        else:
            d[2]=1
        n/=2
    for i in range(3,int(n**0.5)+1,2):
        while(n%i==0):
            if i in d:
                d[i]+=1
            else:
                d[i]=1
            n/=i
    n=int(n)
    if n>1:
        d[n]=1
    return d

#####greatest common divisor of two numbers#####
def gcd(a,b):
    if b==0:
        return a
    return gcd(b,a%b)

#####least common multiplyer of two numbers#####
def lcm(a,b):
    return (a*b)//gcd(a,b)

#####function that return all the letters#####
def alphabates():
    return "abcdefghijklmnopqrstuvwxyz"

#####binary search O(logN)#####
def binary_search(ls,n,flag):
    low=0
    hi=n-1
    while(low<=hi):
        mid=(low+hi)//2
        if ls[mid]==flag:
            return mid
        elif ls[mid]>flag:
            hi=mid-1
        else:
            low=mid+1
    return -1

#####quadratic roots#####
def qdrt(a,b,c):
    chk=b*b-4*a*c
    if chk>=0:
        ans1=(-b+chk**0.5)/(2*a)
        ans2=(-b-chk**0.5)/(2*a)
        return [int(ans1),int(ans2)]
    return -1
#####permutations#####
def permutation(n,r):
    if n<r:
        return 0
    ans=1
    for i in range(n-r+1,n+1):
        ans*=i
    return ans

#####combinations#####
def combination(n,r):
    if n<r:
        return 0
    ans=1
    for i in range(r):
        ans*=(n-i)
    div=1
    for i in range(2,r+1):
        div*=i
    return ans//div

#####check if n is power of 2#####
def ispow2(n):
    ans=int(math.log(n,2))
    if 2**ans==n:
        return True
    return False

#####2d matrix#####
def _2dmat(n):
    flag=[[i+1 for i in range(n)]]
    for i in range(2,n+1):
        new=[]
        for j in range((i-1)*n+1,i*n+1):
            new.append(j)
        flag.append(new)
    return flag


#####sum till nth#####
def sumn(n):
    return (n*(n+1))//2
        
#####taking an array/list as input#####
def arrinp():
    return list(map(int,input().split()))
#####taking multiple inputs#####
def mult_inp():
    return map(int,input().split())
#####Main function starts from here#####
def main():
    n=int(input())
    k=0
    for i in range(2*n,4*n,2):
        if k==n:
            print('\n')
            break
        print(i,end=" ")
        k+=1
    print()
if __name__=='__main__':
    for i in range(int(input())):
        main()