#include <bits/stdc++.h>
using namespace std;
const int N = 200007;
int n, k, e, u, A[N], E[N];
struct W {
  int x, o;
  bool operator<(const W &_) const { return x < _.x; }
} S[N << 1];
int eat(int d) {
  int t = 0, r = 0;
  for (int i = 1; i <= n; ++i) {
    S[++t] = (W){A[i] - d, 1};
    S[++t] = (W){A[i] + d + 1, -1};
  }
  S[++t] = (W){2000000007, 0};
  sort(S + 1, S + 1 + t);
  int x = 0;
  for (int i = 1; i <= t; ++i) {
    x += S[i].o;
    if (S[i].x != S[i + 1].x && x > r) u = S[i].x, r = x;
  }
  return r;
}
void O_O() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &A[i]);
  int l = 0, r = A[n];
  while (l <= r) {
    int m = (l + r) >> 1;
    if (eat(m) > k)
      r = m - 1;
    else
      l = m + 1;
  }
  eat(l);
  printf("%d\n", u);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) O_O();
  return 0;
}

### 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 N = 200007;
int n, k, e, u, A[N], E[N];
struct W {
  int x, o;
  bool operator<(const W &_) const { return x < _.x; }
} S[N << 1];
int eat(int d) {
  int t = 0, r = 0;
  for (int i = 1; i <= n; ++i) {
    S[++t] = (W){A[i] - d, 1};
    S[++t] = (W){A[i] + d + 1, -1};
  }
  S[++t] = (W){2000000007, 0};
  sort(S + 1, S + 1 + t);
  int x = 0;
  for (int i = 1; i <= t; ++i) {
    x += S[i].o;
    if (S[i].x != S[i + 1].x && x > r) u = S[i].x, r = x;
  }
  return r;
}
void O_O() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &A[i]);
  int l = 0, r = A[n];
  while (l <= r) {
    int m = (l + r) >> 1;
    if (eat(m) > k)
      r = m - 1;
    else
      l = m + 1;
  }
  eat(l);
  printf("%d\n", u);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) O_O();
  return 0;
}

```