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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
const double pi = acos(-1.0);
const long long mod = 1000000007;
const long long maxn = 1e6 + 5;
long long power(long long x, unsigned long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res % mod;
}
long long modOp(long long a, long long b) {
  long long c = a % b;
  if (c < 0) c += b;
  return c;
}
long long iv[maxn + 1];
void mod_inv() {
  iv[1] = 1;
  for (long long i = 2; i <= maxn; i++) {
    iv[i] = (mod - mod / i * iv[mod % i] % mod) % mod;
  }
}
long long Logn(long long n, long long base) {
  return (n > base - 1) ? 1 + Logn(n / base, base) : 0;
}
vector<vector<int>> adj;
vector<int> sub_tree_size;
vector<array<int, 2>> edge;
void dfs(int child, int parant) {
  sub_tree_size[child] = 1;
  for (int x : adj[child]) {
    if (x == parant) continue;
    dfs(x, child);
    sub_tree_size[child] += sub_tree_size[x];
  }
}
int isPrime[10000001];
vector<long long> v, w;
vector<vector<long long>> pff(1000000), pr(1000000);
void sieve() {
  for (int i = 2; i * i <= 1000000; i++) {
    if (isPrime[i] == 0) {
      for (int j = i + i; j <= 1000000; j = j + i) {
        isPrime[j] = 1;
      }
    }
  }
  for (int i = 2; i <= 1000000; i++)
    if (isPrime[i] == 0) v.push_back(i);
  return;
}
void run_case() {
  long long n, a;
  cin >> n;
  a = n * 2;
  for (int i = 0; i < n; i++) {
    cout << a << " ";
    a += 2;
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int test;
  cin >> test;
  while (test--) run_case();
  return 0;
}