#include <bits/stdc++.h>
using namespace std;
const long long int INF = 9223372036854775807;
const long long int mod = 998244353;
int MOD(int a, int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
int max3(int a, int b, int c) { return max(c, max(a, b)); }
int min3(int a, int b, int c) { return min(a, min(b, c)); }
long long int power(long long int x, long long int y, long long int mod1) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod1;
    y = y >> 1;
    x = (x * x) % mod1;
  }
  return res;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int abso(long long int x) {
  if (x < 0) {
    return -x;
  }
  return x;
}
long long int ceiling(long long int a, long long int b) {
  if (a % b == 0) {
    return a / b;
  } else {
    return a / b + 1;
  }
}
long long int modinv(long long int x, long long int p) {
  return power(x, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<long long int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long int ans = 0;
    long long int dist = INF;
    for (int i = 0; i + k < n; i++) {
      if ((a[i + k] - a[i] + 1) / 2 < dist) {
        ans = (a[i] + a[i + k]) / 2;
        dist = (a[i + k] - a[i] + 1) / 2;
      }
    }
    cout << ans << endl;
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 long long int INF = 9223372036854775807;
const long long int mod = 998244353;
int MOD(int a, int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
int max3(int a, int b, int c) { return max(c, max(a, b)); }
int min3(int a, int b, int c) { return min(a, min(b, c)); }
long long int power(long long int x, long long int y, long long int mod1) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod1;
    y = y >> 1;
    x = (x * x) % mod1;
  }
  return res;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int abso(long long int x) {
  if (x < 0) {
    return -x;
  }
  return x;
}
long long int ceiling(long long int a, long long int b) {
  if (a % b == 0) {
    return a / b;
  } else {
    return a / b + 1;
  }
}
long long int modinv(long long int x, long long int p) {
  return power(x, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<long long int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long int ans = 0;
    long long int dist = INF;
    for (int i = 0; i + k < n; i++) {
      if ((a[i + k] - a[i] + 1) / 2 < dist) {
        ans = (a[i] + a[i + k]) / 2;
        dist = (a[i + k] - a[i] + 1) / 2;
      }
    }
    cout << ans << endl;
  }
}

```