#include <bits/stdc++.h>
using namespace std;
vector<bool> visited(300000, false);
vector<vector<int>> v;
vector<int> distanc(300000, 0);
queue<int> q;
struct sort_pred {
  bool operator()(const std::pair<int, long long> &left,
                  const std::pair<int, long long> &right) {
    return left.second < right.second;
  }
};
char asciitolower(char in) {
  if (in <= 'Z' && in >= 'A') return in - ('Z' - 'z');
  return in;
}
vector<int> vdeep_copy(vector<int> a) {
  vector<int> b(a.size());
  for (int i = 0; i < a.size(); i++) {
    b[i] = a[i];
  }
  return b;
}
int arr_max(vector<int> a) {
  int ans = -1000000000;
  for (int i = 0; i < a.size(); i++) ans = max(a[i], ans);
  return ans;
}
void to_lower(string &st) {
  std::transform(st.begin(), st.end(), st.begin(), asciitolower);
}
void dfs(int node) {
  if (visited[node]) return;
  visited[node] = true;
  for (auto u : v[node]) dfs(u);
}
void bfs(int x) {
  visited[x] = true;
  distanc[x] = 0;
  q.push(x);
  while (!q.empty()) {
    int s = q.front();
    q.pop();
    for (auto u : v[s]) {
      if (visited[u]) continue;
      visited[u] = true;
      distanc[u] = distanc[s] + 1;
      q.push(u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  vector<long long> reqs(t);
  for (int q = 0; q < t; q++) {
    int n, m;
    cin >> n >> m;
    int k = n / 2 + 1;
    int g = n % 2 ? n / 2 : (n + 1) / 2;
    vector<long long> a(n * m);
    for (int i = 0; i < n * m; i++) cin >> a[i];
    long long sm = 0;
    for (int i = 0; i < m; i++) {
      sm += a[n * m - 1 - g - k * i];
    }
    reqs[q] = sm;
  }
  for (int i = 0; i < t; i++) cout << reqs[i] << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

A median of an array of integers of length n is the number standing on the ⌈ {n/2} ⌉ (rounding up) position in the non-decreasing ordering of its elements. Positions are numbered starting with 1. For example, a median of the array [2, 6, 4, 1, 3, 5] is equal to 3. There exist some other definitions of the median, but in this problem, we will use the described one.

Given two integers n and k and non-decreasing array of nk integers. Divide all numbers into k arrays of size n, such that each number belongs to exactly one array.

You want the sum of medians of all k arrays to be the maximum possible. Find this maximum possible sum.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The next 2t lines contain descriptions of test cases.

The first line of the description of each test case contains two integers n, k (1 ≤ n, k ≤ 1000).

The second line of the description of each test case contains nk integers a_1, a_2, …, a_{nk} (0 ≤ a_i ≤ 10^9) — given array. It is guaranteed that the array is non-decreasing: a_1 ≤ a_2 ≤ … ≤ a_{nk}.

It is guaranteed that the sum of nk for all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case print a single integer — the maximum possible sum of medians of all k arrays.

Example

Input


6
2 4
0 24 34 58 62 64 69 78
2 2
27 61 81 91
4 3
2 4 16 18 21 27 36 53 82 91 92 95
3 4
3 11 12 22 33 35 38 67 69 71 94 99
2 1
11 41
3 3
1 1 1 1 1 1 1 1 1


Output


165
108
145
234
11
3

Note

The examples of possible divisions into arrays for all test cases of the first test:

Test case 1: [0, 24], [34, 58], [62, 64], [69, 78]. The medians are 0, 34, 62, 69. Their sum is 165.

Test case 2: [27, 61], [81, 91]. The medians are 27, 81. Their sum is 108.

Test case 3: [2, 91, 92, 95], [4, 36, 53, 82], [16, 18, 21, 27]. The medians are 91, 36, 18. Their sum is 145.

Test case 4: [3, 33, 35], [11, 94, 99], [12, 38, 67], [22, 69, 71]. The medians are 33, 94, 38, 69. Their sum is 234.

Test case 5: [11, 41]. The median is 11. The sum of the only median is 11.

Test case 6: [1, 1, 1], [1, 1, 1], [1, 1, 1]. The medians are 1, 1, 1. Their sum is 3.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<bool> visited(300000, false);
vector<vector<int>> v;
vector<int> distanc(300000, 0);
queue<int> q;
struct sort_pred {
  bool operator()(const std::pair<int, long long> &left,
                  const std::pair<int, long long> &right) {
    return left.second < right.second;
  }
};
char asciitolower(char in) {
  if (in <= 'Z' && in >= 'A') return in - ('Z' - 'z');
  return in;
}
vector<int> vdeep_copy(vector<int> a) {
  vector<int> b(a.size());
  for (int i = 0; i < a.size(); i++) {
    b[i] = a[i];
  }
  return b;
}
int arr_max(vector<int> a) {
  int ans = -1000000000;
  for (int i = 0; i < a.size(); i++) ans = max(a[i], ans);
  return ans;
}
void to_lower(string &st) {
  std::transform(st.begin(), st.end(), st.begin(), asciitolower);
}
void dfs(int node) {
  if (visited[node]) return;
  visited[node] = true;
  for (auto u : v[node]) dfs(u);
}
void bfs(int x) {
  visited[x] = true;
  distanc[x] = 0;
  q.push(x);
  while (!q.empty()) {
    int s = q.front();
    q.pop();
    for (auto u : v[s]) {
      if (visited[u]) continue;
      visited[u] = true;
      distanc[u] = distanc[s] + 1;
      q.push(u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  vector<long long> reqs(t);
  for (int q = 0; q < t; q++) {
    int n, m;
    cin >> n >> m;
    int k = n / 2 + 1;
    int g = n % 2 ? n / 2 : (n + 1) / 2;
    vector<long long> a(n * m);
    for (int i = 0; i < n * m; i++) cin >> a[i];
    long long sm = 0;
    for (int i = 0; i < m; i++) {
      sm += a[n * m - 1 - g - k * i];
    }
    reqs[q] = sm;
  }
  for (int i = 0; i < t; i++) cout << reqs[i] << endl;
}

```