#include <bits/stdc++.h>
using namespace std;
long long n, k, t;
long long ans;
long long a[400050] = {0};
int qs(long long l, long long r) {
  long long i, j, mid, t;
  i = l;
  j = r;
  mid = a[(i + j) / 2];
  while (1) {
    while (a[i] < mid) i++;
    while (a[j] > mid) j--;
    if (i <= j) {
      t = a[i];
      a[i] = a[j];
      a[j] = t;
      i++;
      j--;
    }
    if (i >= j) break;
  }
  if (l < j) qs(l, j);
  if (i < r) qs(i, r);
  return 0;
}
int main() {
  scanf("%lld", &t);
  long long x;
  for (; t > 0; t--) {
    scanf("%d %d", &n, &k);
    int i, j;
    for (i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    ans = 1;
    for (i = 1; i <= n; i++) {
      if (k + i > n) break;
      if (a[k + ans] - a[ans] > a[k + i] - a[i]) {
        ans = i;
      }
    }
    x = (a[k + ans] + a[ans]) / 2;
    printf("%lld\n", x);
  }
}

### 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;
long long n, k, t;
long long ans;
long long a[400050] = {0};
int qs(long long l, long long r) {
  long long i, j, mid, t;
  i = l;
  j = r;
  mid = a[(i + j) / 2];
  while (1) {
    while (a[i] < mid) i++;
    while (a[j] > mid) j--;
    if (i <= j) {
      t = a[i];
      a[i] = a[j];
      a[j] = t;
      i++;
      j--;
    }
    if (i >= j) break;
  }
  if (l < j) qs(l, j);
  if (i < r) qs(i, r);
  return 0;
}
int main() {
  scanf("%lld", &t);
  long long x;
  for (; t > 0; t--) {
    scanf("%d %d", &n, &k);
    int i, j;
    for (i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    ans = 1;
    for (i = 1; i <= n; i++) {
      if (k + i > n) break;
      if (a[k + ans] - a[ans] > a[k + i] - a[i]) {
        ans = i;
      }
    }
    x = (a[k + ans] + a[ans]) / 2;
    printf("%lld\n", x);
  }
}

```