#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
vector<char*> tokenizer(const char* args) {
  char* token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char*> v({token});
  while (token = strtok(NULL, ", ")) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
void debugg(vector<char*> args) { cerr << "\b\b "; }
template <typename Head, typename... Tail>
void debugg(vector<char*> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
long long powm(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
const long long mod = 1e9 + 7;
const long long N = 1000005;
const long long inf = 1e9;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
  };
  sort(a.begin(), a.end());
  long long ans = inf, ind = a[0];
  for (long long i = 0; i < (n - k); i++) {
    if (ans > (a[i + k] - a[i] + 1) / 2)
      ans = (a[i + k] - a[i] + 1) / 2, ind = a[i] + ans;
  }
  cout << ind << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int tc = 1;
  cin >> tc;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
#pragma GCC optimize("Ofast")
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
vector<char*> tokenizer(const char* args) {
  char* token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char*> v({token});
  while (token = strtok(NULL, ", ")) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
void debugg(vector<char*> args) { cerr << "\b\b "; }
template <typename Head, typename... Tail>
void debugg(vector<char*> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
long long powm(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
const long long mod = 1e9 + 7;
const long long N = 1000005;
const long long inf = 1e9;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
  };
  sort(a.begin(), a.end());
  long long ans = inf, ind = a[0];
  for (long long i = 0; i < (n - k); i++) {
    if (ans > (a[i + k] - a[i] + 1) / 2)
      ans = (a[i + k] - a[i] + 1) / 2, ind = a[i] + ans;
  }
  cout << ind << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int tc = 1;
  cin >> tc;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
  return 0;
}

```