#include <bits/stdc++.h>
using namespace std;
long long int IsPrime(long long int n) {
  if (n == 0 || n == 1) return 0;
  if (n % 2 == 0) return (long long int)2;
  if (n % 3 == 0)
    return (long long int)3;
  else {
    for (long long int i = 5; i * i <= n; i += 6) {
      if (n % i == 0) return i;
      if (n % (i + 2) == 0) return (i + 2);
    }
    return n;
  }
}
bool sec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long int binary_exp(long long int a, long long int n) {
  long long int i = 1;
  while (n > 0) {
    if (n % 2 == 1) i = (i * a) % 1000000007;
    a = (a * a) % 1000000007;
    n /= 2;
  }
  return i;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  int t;
  cin >> t;
  while (t--) {
    long long int n, k, i, j, a, b, minx, maxx, m = LLONG_MAX;
    maxx = minx = -1;
    cin >> n >> k;
    vector<long long int> v(n);
    for (i = 0; i < n; i++) {
      cin >> v[i];
    }
    for (i = 0; i < n - k; i++) {
      if (v[i + k] - v[i] < m) {
        m = v[i + k] - v[i];
        minx = v[i];
        maxx = v[i + k];
      }
    }
    cout << (minx + maxx) / 2 << endl;
  }
}

### Prompt

Please create a solution in cpp 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;
long long int IsPrime(long long int n) {
  if (n == 0 || n == 1) return 0;
  if (n % 2 == 0) return (long long int)2;
  if (n % 3 == 0)
    return (long long int)3;
  else {
    for (long long int i = 5; i * i <= n; i += 6) {
      if (n % i == 0) return i;
      if (n % (i + 2) == 0) return (i + 2);
    }
    return n;
  }
}
bool sec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long int binary_exp(long long int a, long long int n) {
  long long int i = 1;
  while (n > 0) {
    if (n % 2 == 1) i = (i * a) % 1000000007;
    a = (a * a) % 1000000007;
    n /= 2;
  }
  return i;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  int t;
  cin >> t;
  while (t--) {
    long long int n, k, i, j, a, b, minx, maxx, m = LLONG_MAX;
    maxx = minx = -1;
    cin >> n >> k;
    vector<long long int> v(n);
    for (i = 0; i < n; i++) {
      cin >> v[i];
    }
    for (i = 0; i < n - k; i++) {
      if (v[i + k] - v[i] < m) {
        m = v[i + k] - v[i];
        minx = v[i];
        maxx = v[i + k];
      }
    }
    cout << (minx + maxx) / 2 << endl;
  }
}

```