#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int t, n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    vector<long long int> point(n);
    for (long long int i = 0; i < n; i++) cin >> point[i];
    if (k == 0) {
      cout << point[0] << endl;
      continue;
    } else if (k == n - 1) {
      cout << (point[0] + point[n - 1]) / 2 << endl;
      continue;
    } else {
      vector<long long int> gaps(n - 1);
      for (long long int i = 1; i < n; i++)
        gaps[i - 1] = point[i] - point[i - 1];
      long long int sum = 0;
      long long int start = 0;
      long long int ind = 0;
      long long int temp = k;
      while (temp--) sum += gaps[ind], ind++;
      long long int ans = point[start] + (sum) / 2;
      long long int dist = ceil((sum * 1.0) / 2.0);
      for (long long int i = ind; i < n - 1; i++) {
        sum += gaps[i] - gaps[start];
        start++;
        if (ceil((sum * 1.0) / 2.0) < dist) {
          dist = ceil((sum * 1.0) / 2.0);
          ans = (point[start]) + (sum) / 2;
        }
      }
      cout << ans << endl;
    }
  }
  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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int t, n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    vector<long long int> point(n);
    for (long long int i = 0; i < n; i++) cin >> point[i];
    if (k == 0) {
      cout << point[0] << endl;
      continue;
    } else if (k == n - 1) {
      cout << (point[0] + point[n - 1]) / 2 << endl;
      continue;
    } else {
      vector<long long int> gaps(n - 1);
      for (long long int i = 1; i < n; i++)
        gaps[i - 1] = point[i] - point[i - 1];
      long long int sum = 0;
      long long int start = 0;
      long long int ind = 0;
      long long int temp = k;
      while (temp--) sum += gaps[ind], ind++;
      long long int ans = point[start] + (sum) / 2;
      long long int dist = ceil((sum * 1.0) / 2.0);
      for (long long int i = ind; i < n - 1; i++) {
        sum += gaps[i] - gaps[start];
        start++;
        if (ceil((sum * 1.0) / 2.0) < dist) {
          dist = ceil((sum * 1.0) / 2.0);
          ans = (point[start]) + (sum) / 2;
        }
      }
      cout << ans << endl;
    }
  }
  return 0;
}

```