#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const int MAX = 100005;
const int MAX1 = 1005;
const int MAX2 = 105;
const int LEN = 105;
const int BASE = 1000000000;
const double EPS = 1e-7;
const int MOD = 1000000007;
long long pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % MOD;
    b /= 2;
    a = (a * a) % MOD;
  }
  return ans;
}
int solve(int k) {}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ifstream in("");
  ofstream out("");
  clock_t begin = clock();
  int q;
  cin >> q;
  for (int i = (0); i < (q); ++i) {
    int n, k;
    cin >> n >> k;
    int arr[200005];
    for (int j = (0); j < (n); ++j) cin >> arr[j];
    pair<int, int> ans = {0x7fffffff, -1};
    for (int j = (0); j < (n - k); ++j) {
      int d = arr[j + k] - arr[j];
      ans = min(ans, {d, arr[j] + d / 2});
    }
    cout << ans.second << '\n';
  }
  clock_t end = clock();
  double esecs = double(end - begin) / CLOCKS_PER_SEC;
  return 0;
}

### Prompt

Your challenge is to write 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 INF = 1000000000;
const int MAX = 100005;
const int MAX1 = 1005;
const int MAX2 = 105;
const int LEN = 105;
const int BASE = 1000000000;
const double EPS = 1e-7;
const int MOD = 1000000007;
long long pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % MOD;
    b /= 2;
    a = (a * a) % MOD;
  }
  return ans;
}
int solve(int k) {}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ifstream in("");
  ofstream out("");
  clock_t begin = clock();
  int q;
  cin >> q;
  for (int i = (0); i < (q); ++i) {
    int n, k;
    cin >> n >> k;
    int arr[200005];
    for (int j = (0); j < (n); ++j) cin >> arr[j];
    pair<int, int> ans = {0x7fffffff, -1};
    for (int j = (0); j < (n - k); ++j) {
      int d = arr[j + k] - arr[j];
      ans = min(ans, {d, arr[j] + d / 2});
    }
    cout << ans.second << '\n';
  }
  clock_t end = clock();
  double esecs = double(end - begin) / CLOCKS_PER_SEC;
  return 0;
}

```