#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y) {
  long long r = 1, z = x;
  while (y) {
    if (y & 1) r *= z;
    z *= z;
    y = y >> 1;
  }
  return r;
}
long long powerm(long long x, long long y, long long p) {
  long long r = 1;
  while (y) {
    if (y & 1) r = (r * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return r % p;
}
long long modinv(long long x, long long m) { return powerm(x, m - 2, m); }
long long logarithm(long long a, long long b) {
  long long x = 0;
  while (a > 1) {
    x++;
    a /= b;
  }
  return x;
}
long long max3(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long min3(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
void under_rated() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int32_t main() {
  under_rated();
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v;
    for (long long i = 0; i < n; ++i) {
      long long x;
      cin >> x;
      v.push_back(x);
    }
    if (n == 1) {
      cout << v[0] << '\n';
    } else {
      long long mi = 99999999999;
      long long idx = -1;
      for (long long i = 0; i < n - k; ++i) {
        if (mi > (v[i + k] - v[i])) {
          mi = v[i + k] - v[i];
          idx = (v[i] + v[k + i]) / 2;
        }
      }
      cout << idx << '\n';
    }
  }
  return 0;
}

### 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 power(long long x, long long y) {
  long long r = 1, z = x;
  while (y) {
    if (y & 1) r *= z;
    z *= z;
    y = y >> 1;
  }
  return r;
}
long long powerm(long long x, long long y, long long p) {
  long long r = 1;
  while (y) {
    if (y & 1) r = (r * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return r % p;
}
long long modinv(long long x, long long m) { return powerm(x, m - 2, m); }
long long logarithm(long long a, long long b) {
  long long x = 0;
  while (a > 1) {
    x++;
    a /= b;
  }
  return x;
}
long long max3(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long min3(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
void under_rated() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int32_t main() {
  under_rated();
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v;
    for (long long i = 0; i < n; ++i) {
      long long x;
      cin >> x;
      v.push_back(x);
    }
    if (n == 1) {
      cout << v[0] << '\n';
    } else {
      long long mi = 99999999999;
      long long idx = -1;
      for (long long i = 0; i < n - k; ++i) {
        if (mi > (v[i + k] - v[i])) {
          mi = v[i + k] - v[i];
          idx = (v[i] + v[k + i]) / 2;
        }
      }
      cout << idx << '\n';
    }
  }
  return 0;
}

```