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;
long long int const mod = 1e9 + 7;
const long long int N = 1e6 + 3;
const long long int M = 1e5 + 5;
bool sortcol(const vector<long long int>& v1, const vector<long long int>& v2) {
  if (v1[1] == v2[1]) return v1[0] < v2[0];
  return v1[1] < v2[1];
}
bool comp(pair<long long int, long long int> p1,
          pair<long long int, long long int> p2) {
  if (p1.second == p2.second) {
    return p1.first < p2.first;
  }
  return p1.second < p2.second;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int findGCD(long long int arr[], long long int n) {
  long long int result = arr[0];
  for (long long int i = 1; i < n; i++) {
    result = gcd(arr[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b % 2 == 0) {
    return power(a, b / 2) * power(a, b / 2);
  } else
    return power(a, b / 2) * power(a, b / 2) * a;
}
long long int power_bit(long long int a, long long int b, long long int m) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2) {
      ans = (ans * a) % m;
    }
    a = (a * a) % m;
    b = b / 2;
  }
  return ans;
}
long long int multiply_bit(long long int a, long long int b, long long int m) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2) {
      ans = (ans + a) % m;
    }
    a = (a + a) % m;
    b = b / 2;
  }
  return ans - 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  long long int t = 1;
  cin >> t;
  for (long long int i = 0; i < t; i++) {
    long long int n;
    cin >> n;
    for (long long int i = 0; i < n; i++) {
      cout << (n + i) * 2 << " ";
    }
    cout << "\n";
  }
  return 0;
}