#include <bits/stdc++.h>
using namespace std;
long long power(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;
}
long long modInverse(long long a, long long p) { return power(a, p - 2, p); }
long long gcd(long long x, long long y) {
  if (x == 0 || y == 0) {
    return max(y, x);
  }
  return gcd(y % x, x);
}
long long gcdExtended(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
};
void prac() {}
int main() {
  prac();
  long long t;
  scanf("%lld", &t);
  ;
  while (t--) {
    long long n, k;
    scanf("%lld %lld", &(n), &(k));
    vector<long long> a;
    for (long long i = 0; i < n; i++) {
      long long x;
      scanf("%lld", &x);
      a.push_back(x);
    }
    long long ans = 0;
    long long mini = 1e10;
    for (long long i = 0; i < n; i++) {
      if (i + k >= n) break;
      long long ret = a[i + k] - a[i];
      ret /= 2;
      if ((a[i + k] - a[i]) % 2 == 1) {
        ret++;
      }
      if (ret < mini) {
        mini = ret;
        ans = (a[i + k] + a[i]) / 2;
      }
    }
    printf("%lld\n", ans);
  }
}

### 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;
long long power(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;
}
long long modInverse(long long a, long long p) { return power(a, p - 2, p); }
long long gcd(long long x, long long y) {
  if (x == 0 || y == 0) {
    return max(y, x);
  }
  return gcd(y % x, x);
}
long long gcdExtended(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
};
void prac() {}
int main() {
  prac();
  long long t;
  scanf("%lld", &t);
  ;
  while (t--) {
    long long n, k;
    scanf("%lld %lld", &(n), &(k));
    vector<long long> a;
    for (long long i = 0; i < n; i++) {
      long long x;
      scanf("%lld", &x);
      a.push_back(x);
    }
    long long ans = 0;
    long long mini = 1e10;
    for (long long i = 0; i < n; i++) {
      if (i + k >= n) break;
      long long ret = a[i + k] - a[i];
      ret /= 2;
      if ((a[i + k] - a[i]) % 2 == 1) {
        ret++;
      }
      if (ret < mini) {
        mini = ret;
        ans = (a[i + k] + a[i]) / 2;
      }
    }
    printf("%lld\n", ans);
  }
}

```