#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
int check(vector<int>& a, int d, int& x) {
  int n = (int)a.size();
  int l = 0, r = 0;
  int res = 0;
  while (r < n) {
    while (r < n && a[r] - a[l] <= d) {
      if (res < r - l + 1) {
        res = r - l + 1;
        x = l;
      }
      ++r;
    }
    if (r == n) break;
    while (a[r] - a[l] > d) {
      ++l;
    }
  }
  return res;
}
void solve() {
  int n, k;
  scanf("%d%d", &n, &k);
  ++k;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  int l = -1, r = INF, m, x;
  while (r - l > 1) {
    m = (l + r) / 2;
    if (check(a, 2 * m, x) >= k)
      r = m;
    else
      l = m;
  }
  check(a, 2 * r, x);
  printf("%d\n", a[x] + r);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

### Prompt

Your task is to create 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 = 1e9;
int check(vector<int>& a, int d, int& x) {
  int n = (int)a.size();
  int l = 0, r = 0;
  int res = 0;
  while (r < n) {
    while (r < n && a[r] - a[l] <= d) {
      if (res < r - l + 1) {
        res = r - l + 1;
        x = l;
      }
      ++r;
    }
    if (r == n) break;
    while (a[r] - a[l] > d) {
      ++l;
    }
  }
  return res;
}
void solve() {
  int n, k;
  scanf("%d%d", &n, &k);
  ++k;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  int l = -1, r = INF, m, x;
  while (r - l > 1) {
    m = (l + r) / 2;
    if (check(a, 2 * m, x) >= k)
      r = m;
    else
      l = m;
  }
  check(a, 2 * r, x);
  printf("%d\n", a[x] + r);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
}

```