#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << ": " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << "| ";
  __f(comma + 1, args...);
}
const long long N = 2e5 + 10;
const long long mod = 1e9 + 7;
long long n, k;
long long a[N];
pair<bool, long long> check(long long x) {
  vector<pair<long long, long long>> c;
  for (long long i = 1; i <= n; ++i) {
    c.emplace_back(a[i] - x, 0);
    c.emplace_back(a[i] + x, 1);
  }
  sort(c.begin(), c.end());
  long long cnt = 0;
  for (auto i : c) {
    if (i.second == 0)
      ++cnt;
    else
      --cnt;
    if (cnt >= k) return {true, i.first};
  }
  return {false, -1};
}
void bs() {}
void solve() {
  scanf("%lld%lld", &n, &k);
  ++k;
  for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  long long left = 0, right = 1e9 + 1;
  long long ans = -1;
  while (left <= right) {
    long long mid = (left + right) >> 1;
    pair<bool, long long> hell = check(mid);
    if (hell.first) {
      ans = hell.second;
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }
  printf("%lld\n", ans);
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) solve();
  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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << ": " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << "| ";
  __f(comma + 1, args...);
}
const long long N = 2e5 + 10;
const long long mod = 1e9 + 7;
long long n, k;
long long a[N];
pair<bool, long long> check(long long x) {
  vector<pair<long long, long long>> c;
  for (long long i = 1; i <= n; ++i) {
    c.emplace_back(a[i] - x, 0);
    c.emplace_back(a[i] + x, 1);
  }
  sort(c.begin(), c.end());
  long long cnt = 0;
  for (auto i : c) {
    if (i.second == 0)
      ++cnt;
    else
      --cnt;
    if (cnt >= k) return {true, i.first};
  }
  return {false, -1};
}
void bs() {}
void solve() {
  scanf("%lld%lld", &n, &k);
  ++k;
  for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  long long left = 0, right = 1e9 + 1;
  long long ans = -1;
  while (left <= right) {
    long long mid = (left + right) >> 1;
    pair<bool, long long> hell = check(mid);
    if (hell.first) {
      ans = hell.second;
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }
  printf("%lld\n", ans);
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```