#include <bits/stdc++.h>
using namespace std;
int64_t T;
int64_t N;
int64_t K;
vector<int64_t> A;
int main() {
  ios_base::sync_with_stdio(false);
  const unsigned seed = chrono::system_clock::now().time_since_epoch().count();
  mt19937 gen(seed);
  cin >> T;
  for (int64_t t = 0; t < T; ++t) {
    cin >> N >> K;
    A.clear();
    int64_t a;
    for (int64_t n = 0; n < N; ++n) {
      cin >> a;
      A.push_back(a);
    }
    int64_t best_ix = 0;
    int64_t best_range = A[K] - A[0];
    for (int64_t i = 1; i < N - K; ++i) {
      const int64_t candidate_range = A[i + K] - A[i];
      if (candidate_range < best_range) {
        best_range = candidate_range;
        best_ix = i;
      }
    }
    const int64_t ans = (A[best_ix + K] + A[best_ix]) / 2LL;
    cout << ans << "\n";
  }
  cin >> N;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
int64_t T;
int64_t N;
int64_t K;
vector<int64_t> A;
int main() {
  ios_base::sync_with_stdio(false);
  const unsigned seed = chrono::system_clock::now().time_since_epoch().count();
  mt19937 gen(seed);
  cin >> T;
  for (int64_t t = 0; t < T; ++t) {
    cin >> N >> K;
    A.clear();
    int64_t a;
    for (int64_t n = 0; n < N; ++n) {
      cin >> a;
      A.push_back(a);
    }
    int64_t best_ix = 0;
    int64_t best_range = A[K] - A[0];
    for (int64_t i = 1; i < N - K; ++i) {
      const int64_t candidate_range = A[i + K] - A[i];
      if (candidate_range < best_range) {
        best_range = candidate_range;
        best_ix = i;
      }
    }
    const int64_t ans = (A[best_ix + K] + A[best_ix]) / 2LL;
    cout << ans << "\n";
  }
  cin >> N;
  return 0;
}

```