#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const int INF = 2000000000;
int arr[N];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    int res = INF, ans;
    for (int i = 1; i + k <= n; i++) {
      int mid = (arr[i] + arr[i + k]) / 2;
      int d = max(mid - arr[i], arr[i + k] - mid);
      if (d < res) {
        res = d, ans = mid;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

At first, there was a legend related to the name of the problem, but now it's just a formal statement.

You are given n points a_1, a_2, ..., a_n on the OX axis. Now you are asked to find such an integer point x on OX axis that f_k(x) is minimal possible.

The function f_k(x) can be described in the following way: 

  * form a list of distances d_1, d_2, ..., d_n where d_i = |a_i - x| (distance between a_i and x); 
  * sort list d in non-descending order; 
  * take d_{k + 1} as a result. 



If there are multiple optimal answers you can print any of them.

Input

The first line contains single integer T ( 1 ≤ T ≤ 2 ⋅ 10^5) — number of queries. Next 2 ⋅ T lines contain descriptions of queries. All queries are independent. 

The first line of each query contains two integers n, k (1 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ k < n) — the number of points and constant k.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9) — points in ascending order.

It's guaranteed that ∑{n} doesn't exceed 2 ⋅ 10^5.

Output

Print T integers — corresponding points x which have minimal possible value of f_k(x). If there are multiple answers you can print any of them.

Example

Input


3
3 2
1 2 5
2 1
1 1000000000
1 0
4


Output


3
500000000
4

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const int INF = 2000000000;
int arr[N];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    int res = INF, ans;
    for (int i = 1; i + k <= n; i++) {
      int mid = (arr[i] + arr[i + k]) / 2;
      int d = max(mid - arr[i], arr[i + k] - mid);
      if (d < res) {
        res = d, ans = mid;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```