#include <bits/stdc++.h>
using namespace std;
const int N = 212345;
int a[N], d[N];
int n, k;
int tri_div() {
  if (n == 1) return a[0];
  int l = 0, r = k, d = a[r] - a[l];
  for (int i = r + 1; i < n; ++i) {
    int d2 = a[i] - a[i - k];
    if (d2 < d) {
      d = d2;
      r = i;
      l = i - k;
    }
  }
  return (a[r] + a[l]) / 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  scanf("%d", &T);
  for (int ic = 1; ic <= T; ++ic) {
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; ++i) {
      scanf("%d", a + i);
    }
    int x = tri_div();
    printf("%d\n", x);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 = 212345;
int a[N], d[N];
int n, k;
int tri_div() {
  if (n == 1) return a[0];
  int l = 0, r = k, d = a[r] - a[l];
  for (int i = r + 1; i < n; ++i) {
    int d2 = a[i] - a[i - k];
    if (d2 < d) {
      d = d2;
      r = i;
      l = i - k;
    }
  }
  return (a[r] + a[l]) / 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  scanf("%d", &T);
  for (int ic = 1; ic <= T; ++ic) {
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; ++i) {
      scanf("%d", a + i);
    }
    int x = tri_div();
    printf("%d\n", x);
  }
  return 0;
}

```