#include <bits/stdc++.h>
using namespace std;
const int Maxn = 2e5 + 7;
const long long Maxll = 1e18 + 7;
const int inf = 1e9 + 7;
int t;
int n, k;
int aa[Maxn];
long long ans;
long long pre[Maxn];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) scanf("%d", &aa[i]);
    sort(aa + 1, aa + 1 + n);
    long long Minsum = Maxll;
    int dot;
    for (int i = 1; i <= n - k; i++) {
      long long dec = aa[i + k] - aa[i];
      if (dec < Minsum) {
        Minsum = dec;
        dot = i;
      }
    }
    ans = (aa[dot] + aa[dot + k]) / 2;
    printf("%lld\n", ans);
  }
  return 0;
}

### 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 Maxn = 2e5 + 7;
const long long Maxll = 1e18 + 7;
const int inf = 1e9 + 7;
int t;
int n, k;
int aa[Maxn];
long long ans;
long long pre[Maxn];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) scanf("%d", &aa[i]);
    sort(aa + 1, aa + 1 + n);
    long long Minsum = Maxll;
    int dot;
    for (int i = 1; i <= n - k; i++) {
      long long dec = aa[i + k] - aa[i];
      if (dec < Minsum) {
        Minsum = dec;
        dot = i;
      }
    }
    ans = (aa[dot] + aa[dot + k]) / 2;
    printf("%lld\n", ans);
  }
  return 0;
}

```