#include <bits/stdc++.h>
using namespace std;
int T;
int n, k;
int a[200010];
int minDist;
int minPos;
void checkDist(int l, int r, int realPos) {
  int curDistL = realPos - l;
  int curDistR = r - realPos;
  int curDist;
  if (curDistL < curDistR)
    curDist = curDistR;
  else
    curDist = curDistL;
  if ((minDist == -1) || (minDist > curDist)) {
    minDist = curDist;
    minPos = realPos;
  }
}
int main() {
  scanf("%d", &T);
  for (int ii = 0; ii < T; ii++) {
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++) scanf("%d", &(a[i]));
    minDist = -1;
    for (int l = 0; l < n - k; l++) {
      int r = l + k;
      int realNum = (a[l] + a[r]);
      int realPos;
      if (realNum % 2 == 0) {
        realPos = realNum / 2;
        checkDist(a[l], a[r], realPos);
      } else {
        realPos = realNum / 2;
        checkDist(a[l], a[r], realPos);
        checkDist(a[l], a[r], realPos + 1);
      }
    }
    printf("%d\n", minPos);
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int T;
int n, k;
int a[200010];
int minDist;
int minPos;
void checkDist(int l, int r, int realPos) {
  int curDistL = realPos - l;
  int curDistR = r - realPos;
  int curDist;
  if (curDistL < curDistR)
    curDist = curDistR;
  else
    curDist = curDistL;
  if ((minDist == -1) || (minDist > curDist)) {
    minDist = curDist;
    minPos = realPos;
  }
}
int main() {
  scanf("%d", &T);
  for (int ii = 0; ii < T; ii++) {
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++) scanf("%d", &(a[i]));
    minDist = -1;
    for (int l = 0; l < n - k; l++) {
      int r = l + k;
      int realNum = (a[l] + a[r]);
      int realPos;
      if (realNum % 2 == 0) {
        realPos = realNum / 2;
        checkDist(a[l], a[r], realPos);
      } else {
        realPos = realNum / 2;
        checkDist(a[l], a[r], realPos);
        checkDist(a[l], a[r], realPos + 1);
      }
    }
    printf("%d\n", minPos);
  }
  return 0;
}

```