#include <bits/stdc++.h>
using namespace std;
template <class T>
T mceil(T a, T b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
const long long int M = 1e9 + 7;
const long long int N = 2e5 + 5;
const long long int inf = 9e18;
vector<long long int> vis;
vector<vector<long long int> > adj;
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) cin >> v[i];
  if (k == 0) {
    cout << v[0] << "\n";
    return;
  }
  long long int x = -1;
  long long int mn = inf;
  for (long long int i = k; i < n; i++) {
    long long int a = v[i - k];
    long long int b = v[i];
    long long int x1 = (b + a) / 2;
    long long int mx1 = max(x1 - a, b - x1);
    if (mx1 < mn) {
      mn = mx1;
      x = x1;
    }
  }
  cout << x << "\n";
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  for (long long int i = 1; i < t + 1; i++) {
    solve();
  }
}

### Prompt

Please provide a CPP coded solution 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;
template <class T>
T mceil(T a, T b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
const long long int M = 1e9 + 7;
const long long int N = 2e5 + 5;
const long long int inf = 9e18;
vector<long long int> vis;
vector<vector<long long int> > adj;
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) cin >> v[i];
  if (k == 0) {
    cout << v[0] << "\n";
    return;
  }
  long long int x = -1;
  long long int mn = inf;
  for (long long int i = k; i < n; i++) {
    long long int a = v[i - k];
    long long int b = v[i];
    long long int x1 = (b + a) / 2;
    long long int mx1 = max(x1 - a, b - x1);
    if (mx1 < mn) {
      mn = mx1;
      x = x1;
    }
  }
  cout << x << "\n";
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  for (long long int i = 1; i < t + 1; i++) {
    solve();
  }
}

```