#include <bits/stdc++.h>
using namespace std;
void print_width(long long x) {
  std::cout << std::fixed;
  std::cout << std::setprecision(x);
}
long long power(long long x, long long y, long long p = 1000000007) {
  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;
}
void printArr(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
}
void printVector(std::vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
long long Min(long long a, long long b, long long c) {
  if (a < b and a < c) return a;
  if (b < c) return b;
  return c;
}
void initialize(long long arr[], long long n) {
  for (long long i = 0; i <= n; i++) arr[i] = i;
}
long long root(long long arr[], long long i) {
  while (arr[i] != i) {
    arr[i] = arr[arr[i]];
    i = arr[i];
  }
  return i;
}
void Union(long long arr[], long long a, long long b) {
  long long root_a = root(arr, a);
  long long root_b = root(arr, b);
  arr[root_a] = root_b;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long power_wm(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long double dpower(long double x, long long y) {
  long double res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
std::vector<long long> vsum(std::vector<long long> a) {
  std::vector<long long> s(a.size());
  s[0] = a[0];
  for (long long i = 1; i <= a.size() - 1; i++) {
    s[i] = s[i - 1] + a[i];
  }
  return s;
}
int h[100001] = {0};
long long d[100001] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long l;
  cin >> l;
  while (l--) {
    long long n, k;
    cin >> n >> k;
    std::vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    sort(a.begin(), a.end());
    ;
    long long t = k + 1;
    long long x = 1;
    long long d = 99999999999999999;
    for (long long i = 0; i <= n - t; i++) {
      long long a1 = a[i];
      long long a2 = a[i + k];
      if ((a2 - a1) < d) {
        d = a2 - a1;
        x = (a2 + a1) / 2;
      }
    }
    cout << x << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
void print_width(long long x) {
  std::cout << std::fixed;
  std::cout << std::setprecision(x);
}
long long power(long long x, long long y, long long p = 1000000007) {
  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;
}
void printArr(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
}
void printVector(std::vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
long long Min(long long a, long long b, long long c) {
  if (a < b and a < c) return a;
  if (b < c) return b;
  return c;
}
void initialize(long long arr[], long long n) {
  for (long long i = 0; i <= n; i++) arr[i] = i;
}
long long root(long long arr[], long long i) {
  while (arr[i] != i) {
    arr[i] = arr[arr[i]];
    i = arr[i];
  }
  return i;
}
void Union(long long arr[], long long a, long long b) {
  long long root_a = root(arr, a);
  long long root_b = root(arr, b);
  arr[root_a] = root_b;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long power_wm(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long double dpower(long double x, long long y) {
  long double res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
std::vector<long long> vsum(std::vector<long long> a) {
  std::vector<long long> s(a.size());
  s[0] = a[0];
  for (long long i = 1; i <= a.size() - 1; i++) {
    s[i] = s[i - 1] + a[i];
  }
  return s;
}
int h[100001] = {0};
long long d[100001] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long l;
  cin >> l;
  while (l--) {
    long long n, k;
    cin >> n >> k;
    std::vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    sort(a.begin(), a.end());
    ;
    long long t = k + 1;
    long long x = 1;
    long long d = 99999999999999999;
    for (long long i = 0; i <= n - t; i++) {
      long long a1 = a[i];
      long long a2 = a[i + k];
      if ((a2 - a1) < d) {
        d = a2 - a1;
        x = (a2 + a1) / 2;
      }
    }
    cout << x << endl;
  }
  return 0;
}

```