#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 5;
long long T, n, m, wz, ans, a[N];
long long pd(long long mid) {
  long long r = 1, res = 0;
  for (long long i = 1; i <= n; i++) {
    while (r <= n && a[r] - a[i] <= mid * 2) r++;
    res = max(res, r - i);
    if (r - i >= m + 1) wz = (a[i] + a[r - 1]) / 2;
  }
  if (res >= m + 1)
    return 1;
  else
    return 0;
}
signed main() {
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld%lld", &n, &m);
    for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
    long long l = 0, r = 1e9, ans = 1e9;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (pd(mid))
        ans = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    pd(ans);
    printf("%d\n", wz);
  }
  return 0;
}

### 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;
const long long N = 1e6 + 5;
long long T, n, m, wz, ans, a[N];
long long pd(long long mid) {
  long long r = 1, res = 0;
  for (long long i = 1; i <= n; i++) {
    while (r <= n && a[r] - a[i] <= mid * 2) r++;
    res = max(res, r - i);
    if (r - i >= m + 1) wz = (a[i] + a[r - 1]) / 2;
  }
  if (res >= m + 1)
    return 1;
  else
    return 0;
}
signed main() {
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld%lld", &n, &m);
    for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
    long long l = 0, r = 1e9, ans = 1e9;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (pd(mid))
        ans = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    pd(ans);
    printf("%d\n", wz);
  }
  return 0;
}

```