#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int64_t lol;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  int64_t testc = 1;
  cin >> testc;
  while (testc--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    for (auto &i : v) cin >> i;
    if (n == 1) {
      cout << v[0] << "\n";
      continue;
    }
    sort(v.begin(), v.end());
    vector<pair<long long, pair<long long, long long>>> seg;
    for (long long i = 0; i + k < n; i++) {
      seg.push_back(make_pair(v[i + k] - v[i], (make_pair(v[i + k], v[i]))));
    }
    sort(seg.begin(), seg.end());
    k = 0;
    lol = seg[k].second.first + seg[k].second.second;
    lol /= 2;
    cout << lol << "\n";
  }
  return EXIT_SUCCESS;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int64_t lol;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  int64_t testc = 1;
  cin >> testc;
  while (testc--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    for (auto &i : v) cin >> i;
    if (n == 1) {
      cout << v[0] << "\n";
      continue;
    }
    sort(v.begin(), v.end());
    vector<pair<long long, pair<long long, long long>>> seg;
    for (long long i = 0; i + k < n; i++) {
      seg.push_back(make_pair(v[i + k] - v[i], (make_pair(v[i + k], v[i]))));
    }
    sort(seg.begin(), seg.end());
    k = 0;
    lol = seg[k].second.first + seg[k].second.second;
    lol /= 2;
    cout << lol << "\n";
  }
  return EXIT_SUCCESS;
}

```