#include <bits/stdc++.h>
using namespace std;
const long long M = 1e6 + 7;
long long n, a[M], k;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long tt;
  cin >> tt;
  while (tt--) {
    cin >> n >> k;
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long ans1 = INT_MAX, ans2 = -1;
    for (long long i = k; i < n; i++) {
      if (a[i] == a[i - k]) {
        ans2 = a[i];
        break;
      }
      long long t =
          (a[i] * a[i] - a[i - k] * a[i - k]) / (2 * a[i] - 2 * a[i - k]);
      if (a[i] - t < ans1) {
        ans1 = a[i] - t;
        ans2 = t;
      }
    }
    cout << ans2 << "\n";
  }
}

### Prompt

Please formulate 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;
const long long M = 1e6 + 7;
long long n, a[M], k;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long tt;
  cin >> tt;
  while (tt--) {
    cin >> n >> k;
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long ans1 = INT_MAX, ans2 = -1;
    for (long long i = k; i < n; i++) {
      if (a[i] == a[i - k]) {
        ans2 = a[i];
        break;
      }
      long long t =
          (a[i] * a[i] - a[i - k] * a[i - k]) / (2 * a[i] - 2 * a[i - k]);
      if (a[i] - t < ans1) {
        ans1 = a[i] - t;
        ans2 = t;
      }
    }
    cout << ans2 << "\n";
  }
}

```