#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
using namespace std;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long double fi = (sqrtl(5) + 1) / 2;
const long double eps = 1e-8;
vector<long long> a;
long long count(long long x, long long k) {
  vector<long long> w(a.size());
  for (long long i = 0; i < a.size(); i++) w[i] = abs(a[i] - x);
  nth_element(w.begin(), w.begin() + k, w.end());
  return w[k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t, u, n, k, i, l, r, m1, m2;
  cin >> t;
  for ((u) = 0; (u) < (t); (u)++) {
    cin >> n >> k;
    a.resize(n);
    for ((i) = 0; (i) < (n); (i)++) cin >> a[i];
    long long pos = 0;
    for ((i) = 0; (i) < (n); (i)++)
      if (i + k < n && a[i + k] - a[i] < a[pos + k] - a[pos]) pos = i;
    cout << a[pos] + (a[pos + k] - a[pos]) / 2 << '\n';
  }
  return 0;
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
using namespace std;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long double fi = (sqrtl(5) + 1) / 2;
const long double eps = 1e-8;
vector<long long> a;
long long count(long long x, long long k) {
  vector<long long> w(a.size());
  for (long long i = 0; i < a.size(); i++) w[i] = abs(a[i] - x);
  nth_element(w.begin(), w.begin() + k, w.end());
  return w[k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t, u, n, k, i, l, r, m1, m2;
  cin >> t;
  for ((u) = 0; (u) < (t); (u)++) {
    cin >> n >> k;
    a.resize(n);
    for ((i) = 0; (i) < (n); (i)++) cin >> a[i];
    long long pos = 0;
    for ((i) = 0; (i) < (n); (i)++)
      if (i + k < n && a[i + k] - a[i] < a[pos + k] - a[pos]) pos = i;
    cout << a[pos] + (a[pos + k] - a[pos]) / 2 << '\n';
  }
  return 0;
}

```