ECR - 148B - Maximum Sum || Solution in cpp || Question of codeforces

 Question 

You are given an array a1,a2,…,an , where all elements are different.   

You have to perform exactly k operations with it. During each operation, you do exactly one of the following two actions (you choose which to do yourself): find two minimum elements in the array, and delete them;find the maximum element in the array, and delete it.

You have to calculate the maximum possible sum of elements in the resulting array.

Input

The first line contains one integer t (1≤t≤104) — the number of test cases.

Each test case consists of two lines:the first line contains two integers n and k (3≤n≤2⋅105; 1≤k≤99999; 2k<n) — the number of elements and operations, respectively.

the second line contains n integers a1,a2,…,an (1≤ai≤109; all ai  are different) — the elements of the array.Additional constraint on the input: the sum of n does not exceed 2⋅105.

Output

For each test case, print one integer — the maximum possible sum of elements in the resulting array.

Example

input

Copy

6
5 1
2 5 1 10 6
5 2
2 5 1 10 6
3 1
1 2 3
6 1
15 22 12 10 13 11
6 2
15 22 12 10 13 11
5 1
999999996 999999999 999999997 999999998 999999995

output

Copy

21
11
3
62
46
3999999986

Note

In the first testcase, applying the first operation produces the following outcome:

• two minimums are 1$1$ and 2$2$; removing them leaves the array as [5,10,6]$[5,10,6]$, with sum 21$21$;
• a maximum is 10$10$; removing it leaves the array as [2,5,1,6]$[2,5,1,6]$, with sum 14$14$.

21$21$ is the best answer.

In the second testcase, it's optimal to first erase two minimums, then a maximum.

Explanation

Take array arr and n , m 

then sort the array arr

calculate prefix sum as s

then remove the array by rm = s[n-(k-i)] - s[2*i];

and make the answer as ans = max(ans,rm)

print final aswer as cout<<ans<<endl;

Code

#include<bits/stdc++.h>

using namespace std;

#define test     long long T;cin>>T;while(T--)

#define all(x) (x).begin(), (x).end()

#define ll     long long

const int N = 2e5 + 5;

int arr[N];

ll s[N];

void solve()

{

  int n,k; cin>>n>>k;

  for(int i=1;i<=n;i++) cin>>arr[i];

  sort(arr+1,arr+1+n);

  for(int i=1;i<=n;i++){

    s[i] = s[i-1] + arr[i];

  }

  ll ans=0;

  for(int i=0;i<=k;i++){

    ll tp = s[n-(k-i)] - s[2*i];

    ans = max(ans,tp);

  }

  cout<<ans<<endl;

}

signed main() {

    test

    //(if you want to take the more test cases you may uncomment it out)

    solve();

}