#include <bits/stdc++.h>
using namespace std;
const int MAX = 100005;
bool parity(int n) { return !(n & 1); }
bool is2(int n) { return n & (n - 1) == 0; }
template <typename T>
T sqr(T a) {
  return a * a;
}
template <typename T>
T binpow(T n, T s, int base) {
  if (s <= 0) return 1LL;
  if (s % 2 == 0) {
    T b = binpow(n, s / 2, base);
    return (1LL * b * b) % base;
  } else
    return (1LL * binpow(n, s - 1, base) * n) % base;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    vector<int> mas(n + 1);
    for (int i = 1; i <= n; ++i) cin >> mas[i];
    int curMin = 1e9 + 6, curAns = 0, ans = 0;
    for (int i = 1; i + k < mas.size(); i++) {
      curAns = (mas[i] + mas[i + k]) / 2;
      if (max(abs(mas[i] - curAns), abs(mas[i + k] - curAns)) < curMin) {
        curMin = max(abs(mas[i] - curAns), abs(mas[i + k] - curAns));
        ans = curAns;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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;
const int MAX = 100005;
bool parity(int n) { return !(n & 1); }
bool is2(int n) { return n & (n - 1) == 0; }
template <typename T>
T sqr(T a) {
  return a * a;
}
template <typename T>
T binpow(T n, T s, int base) {
  if (s <= 0) return 1LL;
  if (s % 2 == 0) {
    T b = binpow(n, s / 2, base);
    return (1LL * b * b) % base;
  } else
    return (1LL * binpow(n, s - 1, base) * n) % base;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    vector<int> mas(n + 1);
    for (int i = 1; i <= n; ++i) cin >> mas[i];
    int curMin = 1e9 + 6, curAns = 0, ans = 0;
    for (int i = 1; i + k < mas.size(); i++) {
      curAns = (mas[i] + mas[i + k]) / 2;
      if (max(abs(mas[i] - curAns), abs(mas[i + k] - curAns)) < curMin) {
        curMin = max(abs(mas[i] - curAns), abs(mas[i + k] - curAns));
        ans = curAns;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```