Problem:
Score : 400 points

Problem Statement

Your friend gave you a dequeue D as a birthday present.

D is a horizontal cylinder that contains a row of N jewels.

The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.

In the beginning, you have no jewel in your hands.

You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):

Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.

Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.

Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.

Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.

Find the maximum possible sum of the values of jewels in your hands after the operations.

Constraints

All values in input are integers.

1 \leq N \leq 50

1 \leq K \leq 100

-10^7 \leq V_i \leq 10^7

Input

Input is given from Standard Input in the following format:

N K
V_1 V_2 ... V_N

Output

Print the maximum possible sum of the values of jewels in your hands after the operations.

Sample Input 1

6 4
-10 8 2 1 2 6

Sample Output 1

14

After the following sequence of operations, you have two jewels of values 8 and 6 in your hands for a total of 14, which is the maximum result.

Do operation A. You take out the jewel of value -10 from the left end of D.

Do operation B. You take out the jewel of value 6 from the right end of D.

Do operation A. You take out the jewel of value 8 from the left end of D.

Do operation D. You insert the jewel of value -10 to the right end of D.

Sample Input 2

6 4
-6 -100 50 -2 -5 -3

Sample Output 2

44

Sample Input 3

6 3
-6 -100 50 -2 -5 -3

Sample Output 3

0

It is optimal to do no operation.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define endl "\n"
#define all(x) x.begin(),x.end()
#define pb push_back
#define ff first 
#define ss second 
#define pii pair<int,int>
#define vi vector<int>
#define vc vector<char>
#define vii vector<pii>
#define vvi vector<vi>
#define fori(i,n) for(int i=1;i<=n;++i)
#define forn(i,n) for(int i=0;i<n;++i)
#define sz(x) int(x.size())
#define error(x) cout<<#x<<" "<<x<<endl
#define IOS  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)

const int N=100*1000;
const int mod=1e9+7;

int min3(int a,int b,int c){return min(a,min(b,c));}
int max3(int a,int b,int c){return max(a,max(b,c));}
int min4(int a,int b,int c,int d){return min(min(a,b),min(c,d));}
int max4(int a,int b,int c,int d){return max(max(a,b),max(c,d));}

#define int int64_t
//int a[N+2];
int v[100];
int pref[100]={0};
int suff[100]={0};
int32_t  main(){
  IOS;
  int n,KK;cin>>n>>KK;
  fori(i,n)cin>>v[i];
  memset(pref,0,sizeof(pref));
  memset(suff,0,sizeof(suff));
  for(int i=1;i<=n;++i)pref[i]=v[i]+pref[i-1];
  for(int i=n;i>=1;--i)suff[i]=v[i]+suff[i+1];
  //total operations is i+j+k 
  int ans=0;
  int K=KK;
  for(int i=0;i<=K;++i){
    for(int j=0;j<=K;++j){
      for(int k=0;k<=K;++k){
        if(i+j+k<=K && i<n-j+1 ){
                    //cout<<i<<" "<<j<<" "<<k<<" i,j,k\n";
          //prefix sum of i 
          //suffix sum of j 
          //remove k neg elements 
          int pf=pref[i];
          int sf=suff[n-j+1];
          //pick negative least elements 
          priority_queue<int>pq;
          for(int x=1;x<=i;++x){
            if(v[x]<0)pq.push(-v[x]);
          }
          for(int x=n-j+1;x<=n;++x){
            if(v[x]<0)pq.push(-v[x]);
          }
          //pick top k ;
          int kol=k;
          int sum=0;
          while(kol-- && pq.size()>0){
            sum+=pq.top();
            pq.pop();
          }
          sum*=-1;
          int val=pf+sf-sum;
                      //cout<<val<<" val\n";
          ans=max(ans,pf+sf-sum);
        }
      }

    }
  }
  cout<<ans<<endl;

  return 0;
}