#include <bits/stdc++.h>
using namespace std;
inline long long fpow(long long n, long long k, int p = 998244353) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
inline long long inv(long long a, long long p = 998244353) {
  return fpow(a, p - 2, p);
}
inline long long addmod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a + val)) >= p) a -= p;
  }
  return a;
}
inline long long submod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a - val)) < 0) a += p;
  }
  return a;
}
inline long long mult(long long a, long long b, long long p = 998244353) {
  return (long long)a * b % p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    vector<long long> a(n + 5);
    for (long long i = 1; i <= n; i++) cin >> a[i];
    long long ans = LLONG_MAX, idx;
    queue<long long> q;
    for (int i = 1; i <= n; i++) {
      if (q.size() < k) {
        q.push(a[i]);
      } else {
        q.pop();
        q.push(a[i]);
      }
      if (q.size() != k) continue;
      long long dist = q.back() - q.front();
      if (dist < ans) {
        ans = dist;
        idx = q.front() + dist / 2;
      }
    }
    cout << idx << "\n";
  }
}

### 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;
inline long long fpow(long long n, long long k, int p = 998244353) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
inline long long inv(long long a, long long p = 998244353) {
  return fpow(a, p - 2, p);
}
inline long long addmod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a + val)) >= p) a -= p;
  }
  return a;
}
inline long long submod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a - val)) < 0) a += p;
  }
  return a;
}
inline long long mult(long long a, long long b, long long p = 998244353) {
  return (long long)a * b % p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    vector<long long> a(n + 5);
    for (long long i = 1; i <= n; i++) cin >> a[i];
    long long ans = LLONG_MAX, idx;
    queue<long long> q;
    for (int i = 1; i <= n; i++) {
      if (q.size() < k) {
        q.push(a[i]);
      } else {
        q.pop();
        q.push(a[i]);
      }
      if (q.size() != k) continue;
      long long dist = q.back() - q.front();
      if (dist < ans) {
        ans = dist;
        idx = q.front() + dist / 2;
      }
    }
    cout << idx << "\n";
  }
}

```