#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
long long po(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool sec(const pair<long long, long long> &a,
         const pair<long long, long long> &b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first > b.first;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, k, mi = 1e18, j, i;
    cin >> n >> k;
    long long a[n];
    for (i = 0; i < n; i++) cin >> a[i];
    for (i = 0; i < n - k; i++)
      if (a[i + k] - a[i] < mi) mi = a[i + k] - a[i], j = i;
    cout << (a[j] + a[j + k]) / 2 << endl;
  }
}

### Prompt

Create a solution in CPP for 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;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
long long po(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool sec(const pair<long long, long long> &a,
         const pair<long long, long long> &b) {
  if (a.first == b.first) return a.second > b.second;
  return a.first > b.first;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, k, mi = 1e18, j, i;
    cin >> n >> k;
    long long a[n];
    for (i = 0; i < n; i++) cin >> a[i];
    for (i = 0; i < n - k; i++)
      if (a[i + k] - a[i] < mi) mi = a[i + k] - a[i], j = i;
    cout << (a[j] + a[j + k]) / 2 << endl;
  }
}

```