#include <bits/stdc++.h>
const int MAXN = 2e5 + 5;
int a[MAXN], T, n, k;
inline void Solve() {
  scanf("%d%d", &n, &k);
  for (register int i = 1; i <= n; ++i) scanf("%d", a + i);
  int ans = k + 1;
  for (register int i = 1; i <= n; ++i) {
    if (i > k && a[i] - a[i - k] < a[ans] - a[ans - k]) {
      ans = i;
    }
  }
  printf("%d\n", (a[ans] + a[ans - k]) / 2);
}
int main() {
  scanf("%d", &T);
  while (T--) Solve();
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
const int MAXN = 2e5 + 5;
int a[MAXN], T, n, k;
inline void Solve() {
  scanf("%d%d", &n, &k);
  for (register int i = 1; i <= n; ++i) scanf("%d", a + i);
  int ans = k + 1;
  for (register int i = 1; i <= n; ++i) {
    if (i > k && a[i] - a[i - k] < a[ans] - a[ans - k]) {
      ans = i;
    }
  }
  printf("%d\n", (a[ans] + a[ans - k]) / 2);
}
int main() {
  scanf("%d", &T);
  while (T--) Solve();
  return 0;
}

```