#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long n, k;
    cin >> n >> k;
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long mi = 1000000000000, mini = 1000000000000, ans;
    for (long long i = 0; i < n - k; i++) {
      long long mid = (a[i] + a[i + k]) / 2;
      mi = max(abs(a[i] - mid), abs(mid - a[i + k]));
      if (mi < mini) {
        ans = mid;
        mini = mi;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long n, k;
    cin >> n >> k;
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long mi = 1000000000000, mini = 1000000000000, ans;
    for (long long i = 0; i < n - k; i++) {
      long long mid = (a[i] + a[i + k]) / 2;
      mi = max(abs(a[i] - mid), abs(mid - a[i + k]));
      if (mi < mini) {
        ans = mid;
        mini = mi;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```