#include <bits/stdc++.h>
using namespace std;
long long int MOD = 1000000007;
long long int inf = 1e15;
void scan(int &x);
long long int powermod(long long int _a, long long int _b, long long int _m) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
long long int string_to_number(string s) {
  long long int x = 0;
  stringstream convert(s);
  convert >> x;
  return x;
}
long long int add(long long int a, long long int b) {
  long long int x = (a + b) % MOD;
  return x;
}
long long int mul(long long int a, long long int b) {
  long long int x = (a * b) % MOD;
  return x;
}
long long int sub(long long int a, long long int b) {
  long long int x = (a - b + MOD) % MOD;
  return x;
}
int possible(int ans, int n, int k, int arr[]) {
  vector<pair<int, int> > events;
  set<int> maxR;
  int intersection = 0, result;
  for (int i = 0; i < n; i++) {
    int tempL = arr[i] - ans;
    int tempR = arr[i] + ans;
    while (!maxR.empty() && tempL > *(maxR.begin())) {
      maxR.erase(maxR.begin());
    }
    maxR.insert(tempR);
    if (maxR.size() > intersection) {
      intersection = maxR.size();
      result = *(maxR.begin());
    }
  }
  if (intersection >= k) {
    return result;
  }
  return -1;
}
void solve() {
  int n, k;
  scan(n);
  scan(k);
  int arr[n];
  k++;
  for (int i = 0; i < n; i++) {
    scan(arr[i]);
  }
  int low = 0, high = 2e9, ans = 0;
  while (low <= high) {
    int mid = (low + high) >> 1;
    int possibleAns = possible(mid, n, k, arr);
    if (possibleAns != -1) {
      ans = possibleAns;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  cout << ans << "\n";
}
int main(void) {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
void scan(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}

### Prompt

Please provide a Cpp coded solution 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>
using namespace std;
long long int MOD = 1000000007;
long long int inf = 1e15;
void scan(int &x);
long long int powermod(long long int _a, long long int _b, long long int _m) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
long long int string_to_number(string s) {
  long long int x = 0;
  stringstream convert(s);
  convert >> x;
  return x;
}
long long int add(long long int a, long long int b) {
  long long int x = (a + b) % MOD;
  return x;
}
long long int mul(long long int a, long long int b) {
  long long int x = (a * b) % MOD;
  return x;
}
long long int sub(long long int a, long long int b) {
  long long int x = (a - b + MOD) % MOD;
  return x;
}
int possible(int ans, int n, int k, int arr[]) {
  vector<pair<int, int> > events;
  set<int> maxR;
  int intersection = 0, result;
  for (int i = 0; i < n; i++) {
    int tempL = arr[i] - ans;
    int tempR = arr[i] + ans;
    while (!maxR.empty() && tempL > *(maxR.begin())) {
      maxR.erase(maxR.begin());
    }
    maxR.insert(tempR);
    if (maxR.size() > intersection) {
      intersection = maxR.size();
      result = *(maxR.begin());
    }
  }
  if (intersection >= k) {
    return result;
  }
  return -1;
}
void solve() {
  int n, k;
  scan(n);
  scan(k);
  int arr[n];
  k++;
  for (int i = 0; i < n; i++) {
    scan(arr[i]);
  }
  int low = 0, high = 2e9, ans = 0;
  while (low <= high) {
    int mid = (low + high) >> 1;
    int possibleAns = possible(mid, n, k, arr);
    if (possibleAns != -1) {
      ans = possibleAns;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  cout << ans << "\n";
}
int main(void) {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
void scan(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}

```