#include <bits/stdc++.h>
using namespace std;
inline void fileio(const char* in, const char* out) {
  freopen(in, "r", stdin);
  freopen(out, "w", stdout);
}
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> VI;
typedef vector<bool> VB;
typedef vector<ll> VL;
const char nl = '\n';
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fll;
const ld EPS = 1e-9, PI = acos(-1);
void deal(int T) {
  int n, k;
  cin >> n >> k;
  VI nums(n);
  for (int i = 0; i < (n); ++i) {
    cin >> nums[i];
  }
  ll best = INT_MAX;
  ll x;
  for (int i = 0, j = k; j < n; ++i, ++j) {
    ll _x = nums[i] + nums[j] >> 1;
    ll dist = max(_x - nums[i], nums[j] - _x);
    if (dist < best) {
      best = dist;
      x = _x;
    }
  }
  cout << x << nl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  for (int i = 0; i < (T); ++i) deal(i);
  cerr << "Time: " << (ld)clock() / CLOCKS_PER_SEC << nl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
inline void fileio(const char* in, const char* out) {
  freopen(in, "r", stdin);
  freopen(out, "w", stdout);
}
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> VI;
typedef vector<bool> VB;
typedef vector<ll> VL;
const char nl = '\n';
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fll;
const ld EPS = 1e-9, PI = acos(-1);
void deal(int T) {
  int n, k;
  cin >> n >> k;
  VI nums(n);
  for (int i = 0; i < (n); ++i) {
    cin >> nums[i];
  }
  ll best = INT_MAX;
  ll x;
  for (int i = 0, j = k; j < n; ++i, ++j) {
    ll _x = nums[i] + nums[j] >> 1;
    ll dist = max(_x - nums[i], nums[j] - _x);
    if (dist < best) {
      best = dist;
      x = _x;
    }
  }
  cout << x << nl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  for (int i = 0; i < (T); ++i) deal(i);
  cerr << "Time: " << (ld)clock() / CLOCKS_PER_SEC << nl;
  return 0;
}

```