#include <bits/stdc++.h>
const int INF = (int)1e9 + 7;
int f(const std::vector<int> &arr, int k, int d) {
  std::list<std::pair<int, int>> events, se;
  for (auto &it : arr) {
    events.push_back({it - d, 0});
    se.push_back({it + d, 1});
  }
  events.merge(se);
  int bal = 0;
  for (auto it : events) {
    bal += (it.second ? -1 : +1);
    if (bal > k) {
      return it.first;
    }
  }
  return -INF;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  int t;
  std::cin >> t;
  while (t--) {
    int n, k;
    std::cin >> n >> k;
    std::vector<int> arr(n);
    for (auto &it : arr) {
      std::cin >> it;
    }
    std::sort((arr).begin(), (arr).end());
    int low = -1, high = (int)1e9;
    while (high - low > 1) {
      int mid = (low + high) / 2;
      if (f(arr, k, mid) > -INF) {
        high = mid;
      } else {
        low = mid;
      }
    }
    std::cout << f(arr, k, high) << "\n";
  }
  return 0;
}

### Prompt

Your task is to create 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>
const int INF = (int)1e9 + 7;
int f(const std::vector<int> &arr, int k, int d) {
  std::list<std::pair<int, int>> events, se;
  for (auto &it : arr) {
    events.push_back({it - d, 0});
    se.push_back({it + d, 1});
  }
  events.merge(se);
  int bal = 0;
  for (auto it : events) {
    bal += (it.second ? -1 : +1);
    if (bal > k) {
      return it.first;
    }
  }
  return -INF;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  int t;
  std::cin >> t;
  while (t--) {
    int n, k;
    std::cin >> n >> k;
    std::vector<int> arr(n);
    for (auto &it : arr) {
      std::cin >> it;
    }
    std::sort((arr).begin(), (arr).end());
    int low = -1, high = (int)1e9;
    while (high - low > 1) {
      int mid = (low + high) / 2;
      if (f(arr, k, mid) > -INF) {
        high = mid;
      } else {
        low = mid;
      }
    }
    std::cout << f(arr, k, high) << "\n";
  }
  return 0;
}

```