#include <bits/stdc++.h>
using namespace std;
vector<vector<long long>> adj;
map<long, bool> vis, viss;
vector<long long> rnk, parent, sz;
int spf[1000000 + 1];
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(nullptr);
  long long t, temp;
  long long n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    vector<long long> v;
    for (long long i = 0; i < n; i++) {
      cin >> temp;
      v.push_back(temp);
    }
    long long yg = k + 1;
    long long mn = LLONG_MAX;
    long long ans = -1;
    if (n == 1) {
      cout << v[0] << "\n";
      continue;
    }
    for (long long i = 0; i + yg - 1 < n; i++) {
      if (mn > (v[i - 1 + yg] - v[i] + 1) / 2) {
        mn = (v[i - 1 + yg] - v[i] + 1) / 2;
        ans = (v[i - 1 + yg] + v[i] + 1) / 2;
      };
    }
    cout << ans << "\n";
  }
  return 0;
}

### 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;
vector<vector<long long>> adj;
map<long, bool> vis, viss;
vector<long long> rnk, parent, sz;
int spf[1000000 + 1];
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(nullptr);
  long long t, temp;
  long long n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    vector<long long> v;
    for (long long i = 0; i < n; i++) {
      cin >> temp;
      v.push_back(temp);
    }
    long long yg = k + 1;
    long long mn = LLONG_MAX;
    long long ans = -1;
    if (n == 1) {
      cout << v[0] << "\n";
      continue;
    }
    for (long long i = 0; i + yg - 1 < n; i++) {
      if (mn > (v[i - 1 + yg] - v[i] + 1) / 2) {
        mn = (v[i - 1 + yg] - v[i] + 1) / 2;
        ans = (v[i - 1 + yg] + v[i] + 1) / 2;
      };
    }
    cout << ans << "\n";
  }
  return 0;
}

```