#include <bits/stdc++.h>
using namespace std;
long long INF = 1000000007;
long long INFLL = (long long)INF * (long long)INF;
long double EPS = 10e-9;
long double pi = 2 * acos(0.0);
long long mod = 1e9 + 7;
long long sum = 0ll;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  long long strt = 0;
  long long nd = strt + k;
  long long dis = INFLL;
  long long ans = a[0];
  while (nd < n) {
    long long mid = a[strt] + a[nd];
    mid /= 2;
    long long maxi = max(abs(mid - a[strt]), abs(mid - a[nd]));
    if (maxi < dis) {
      dis = maxi;
      ans = mid;
    }
    strt++;
    nd++;
  }
  cout << ans << "\n";
}
int32_t main() {
  clock_t start, end;
  start = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) solve();
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
  cerr << "\nTime taken by program is : " << fixed << time_taken
       << setprecision(5);
  cerr << " sec "
       << "\n";
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long INF = 1000000007;
long long INFLL = (long long)INF * (long long)INF;
long double EPS = 10e-9;
long double pi = 2 * acos(0.0);
long long mod = 1e9 + 7;
long long sum = 0ll;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  long long strt = 0;
  long long nd = strt + k;
  long long dis = INFLL;
  long long ans = a[0];
  while (nd < n) {
    long long mid = a[strt] + a[nd];
    mid /= 2;
    long long maxi = max(abs(mid - a[strt]), abs(mid - a[nd]));
    if (maxi < dis) {
      dis = maxi;
      ans = mid;
    }
    strt++;
    nd++;
  }
  cout << ans << "\n";
}
int32_t main() {
  clock_t start, end;
  start = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) solve();
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
  cerr << "\nTime taken by program is : " << fixed << time_taken
       << setprecision(5);
  cerr << " sec "
       << "\n";
  return 0;
}

```