#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6 + 5;
const long long inf = 1e10;
inline int read() {
  int x = 0, w = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') w = 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return w ? x : -x;
}
int ai[maxn];
int main() {
  int t = read();
  while (t--) {
    int n = read(), k = read();
    for (int a = 1; a <= n; a++) ai[a] = read();
    if (n == 1) {
      printf("%d\n", ai[1]);
      continue;
    }
    sort(ai + 1, ai + n + 1);
    long long ans = inf, ans1;
    for (int i = k + 1; i <= n; i++) {
      int j = i - k;
      long long dis = ai[i] + ai[j];
      long long xi = dis / 2;
      if (ans > max(ai[i] - xi, xi - ai[i]))
        ans1 = xi, ans = max(ai[i] - xi, xi - ai[i]);
    }
    printf("%lld\n", ans1);
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 2e6 + 5;
const long long inf = 1e10;
inline int read() {
  int x = 0, w = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') w = 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return w ? x : -x;
}
int ai[maxn];
int main() {
  int t = read();
  while (t--) {
    int n = read(), k = read();
    for (int a = 1; a <= n; a++) ai[a] = read();
    if (n == 1) {
      printf("%d\n", ai[1]);
      continue;
    }
    sort(ai + 1, ai + n + 1);
    long long ans = inf, ans1;
    for (int i = k + 1; i <= n; i++) {
      int j = i - k;
      long long dis = ai[i] + ai[j];
      long long xi = dis / 2;
      if (ans > max(ai[i] - xi, xi - ai[i]))
        ans1 = xi, ans = max(ai[i] - xi, xi - ai[i]);
    }
    printf("%lld\n", ans1);
  }
}

```