#include <bits/stdc++.h>
using namespace std;
int n, k;
int dat[200100];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; i++) scanf("%d", dat + i);
    int mipos = (dat[0] + dat[k]) >> 1, mi = dat[k] - mipos;
    for (int i = 1; i + k < n; i++) {
      int cur = (dat[i] + dat[i + k]) >> 1;
      if (dat[i + k] - cur < mi) {
        mi = dat[i + k] - cur;
        mipos = cur;
      }
    }
    printf("%d\n", mipos);
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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 n, k;
int dat[200100];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; i++) scanf("%d", dat + i);
    int mipos = (dat[0] + dat[k]) >> 1, mi = dat[k] - mipos;
    for (int i = 1; i + k < n; i++) {
      int cur = (dat[i] + dat[i + k]) >> 1;
      if (dat[i + k] - cur < mi) {
        mi = dat[i + k] - cur;
        mipos = cur;
      }
    }
    printf("%d\n", mipos);
  }
  return 0;
}

```