#include <bits/stdc++.h>
using namespace std;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
bool ok = 0;
long long val = pow(2, 32) - 1;
long long fastpow(long long x, long long p) {
  if (!p) return 1;
  long long sq = fastpow(x, p / 2);
  if (sq > val) {
    ok = 1;
    return 0;
  }
  sq *= sq;
  if (sq > val) {
    ok = 1;
    return 0;
  }
  if (p & 1) sq *= x;
  if (sq > val) {
    ok = 1;
    return 0;
  }
  return sq;
}
int main() {
  file();
  fast();
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<int> v1(n);
    for (auto &x : v1) cin >> x;
    long long mn = INT_MAX;
    int ans;
    for (int i = 0; i + k < n; i++) {
      if (v1[i + k] - v1[i] < mn) {
        mn = v1[i + k] - v1[i];
        ans = (v1[i + k] + v1[i]) / 2;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Generate 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;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
bool ok = 0;
long long val = pow(2, 32) - 1;
long long fastpow(long long x, long long p) {
  if (!p) return 1;
  long long sq = fastpow(x, p / 2);
  if (sq > val) {
    ok = 1;
    return 0;
  }
  sq *= sq;
  if (sq > val) {
    ok = 1;
    return 0;
  }
  if (p & 1) sq *= x;
  if (sq > val) {
    ok = 1;
    return 0;
  }
  return sq;
}
int main() {
  file();
  fast();
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<int> v1(n);
    for (auto &x : v1) cin >> x;
    long long mn = INT_MAX;
    int ans;
    for (int i = 0; i + k < n; i++) {
      if (v1[i + k] - v1[i] < mn) {
        mn = v1[i + k] - v1[i];
        ans = (v1[i + k] + v1[i]) / 2;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

```