#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 2e5 + 5;
const int mod = 1e9 + 7;
const long long INF = 2e9 + 7;
long long a[MAX_N];
void solve() {
  int t;
  cin >> t;
  while (t--) {
    int n, k, ans, min_diff = 1e9 + 7;
    cin >> n >> k;
    sort(a, a + n);
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n - k; i++) {
      int mid = (a[i] + a[i + k]) / 2;
      if (min_diff > a[i + k] - a[i]) min_diff = a[i + k] - a[i], ans = mid;
    }
    cout << ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t start = clock();
  solve();
  fprintf(stderr, "Total Time: %.4f\n",
          (double)(clock() - start) / CLOCKS_PER_SEC),
      fflush(stderr);
  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>
using namespace std;
const int MAX_N = 2e5 + 5;
const int mod = 1e9 + 7;
const long long INF = 2e9 + 7;
long long a[MAX_N];
void solve() {
  int t;
  cin >> t;
  while (t--) {
    int n, k, ans, min_diff = 1e9 + 7;
    cin >> n >> k;
    sort(a, a + n);
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n - k; i++) {
      int mid = (a[i] + a[i + k]) / 2;
      if (min_diff > a[i + k] - a[i]) min_diff = a[i + k] - a[i], ans = mid;
    }
    cout << ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t start = clock();
  solve();
  fprintf(stderr, "Total Time: %.4f\n",
          (double)(clock() - start) / CLOCKS_PER_SEC),
      fflush(stderr);
  return 0;
}

```