Codeforces Round #684 (Div. 2) B. Sum of Medians

A median of an array of integers of length n is the number standing on the ⌈n2⌉ (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 a1,a2,…,ank (0≤ai≤109) — given array. It is guaranteed that the array is non-decreasing: a1≤a2≤…≤ank.

It is guaranteed that the sum of nk for all test cases does not exceed 2⋅105.

Output

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

Example

inputCopy

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

outputCopy

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.

倒着去取中位數，看規律要快

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <queue>
#include <functional>
#include <vector>
#include <stack>
#include <set>
#define int long long
using namespace std;
typedef long long ll;
const int maxn=2e5+50;
const int inf=0x3f3f3f3f;
const int MOD=1e9+7;
const int HASH=131;

int a[maxn];
bool cmp(int a,int b)
{
    return a>b;
}
signed main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,k;
        cin>>n>>k;
        for(int i=1;i<=n*k;i++)
        {
            cin>>a[i];
        }
        sort(a+1,a+1+n*k,cmp);
        int ans=0;
        for(int i=(n)/2+1,j=1;j<=k;i+=(n)/2+1,j++)
        {
                ans+=a[i];
        }
        cout<<ans<<endl;
    }
}