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)

// #Sazaの1日1AC
#include<bits/stdc++.h>
using namespace std;
const long long mod=1000000000+7;
const long long INF=1e16;

//マクロ
//型エイリアス
using ll = long long;
using P=pair<long long,long long>;
using vl=vector<long long>;
//ショートカット
#define rep(i,n) for(int i=0;i<n;i++)
#define rep2(i,k,n) for(int i=k;i<n;i++) //半開区間
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
//入力
#define vin(v,N) for(long long i=0;i<N;i++) cin>>v.at(i)
#define lin(n) long long n;cin>>n
#define vlin(v,N) vector<long long>v(N);for(long long i=0;i<N;i++)cin>>v.at(i)

#define max(a,b) max((ll)a,(ll)b)
#define min(a,b) min((ll)a,(ll)b)

//関数
//最大公約数
long long gcd(long long m, long long n){
  long long a=max(m,n);
  long long b=min(m,n);
  long long r=a%b;
  while(r!=0){
    a=b; b=r; r=a%b;
  }
  return b;
}
// 最小公倍数
long long lcd(long long m, long long n){
  return m*n/gcd(m,n);
}
//xのn乗
long long power(long long x, long long N){
  long long ret=1;
  for(long long i=0;i<N;i++)   
    ret*=x;
  return ret;
}
//繰り返し二乗法
long long repeat_squaring(long long x, long long n){
  if(n==0) return 1;
  else if(n%2==0){
    long long t=repeat_squaring(x,n/2);
    return t*t%mod;
  }
  else
    return x*repeat_squaring(x,n-1);
}
//nCr
long long Comb (long long n, long long r);
//nCr (mod)
long long modC(long long n, long long r) {
    vector<long long> f(n+1),inv(n+1),invf(n+1);
    f[0]=f[1]=1;
    invf[0]=invf[1]=1;
    inv[1]=1;
    for (long long i=2;i<n+1;i++) {
        f[i]=(i*f[i-1])%mod;
        inv[i]=mod-((mod/i)*inv[mod%i])%mod;
        invf[i]=(inv[i]*invf[i-1])%mod;
    }
    long long cmb=f[n]*(invf[r]*invf[n-r]%mod)%mod;
    if (n>=r) return cmb;
    else return 0;
}
//素因数分解(mapで返す)
map<long long,long long> factor(long long p){
  ll p1=p;
  map<long long,long long>ret;
  for(long long i=2;i*i<=p1;i++){
     while(p%i==0){
      ret[i]++;
      p/=i;
    }
  }
  if(p!=1)
  ret[p]++;
  return ret;
}
//素数判定
bool is_prime(long long N){
  for(long long i=2;i*i<=N;i++){
    if(N%i==0)
      return false;
  }
  return true;
}
//最大値更新
void chmax(long long &a,long long b){
  a=max(a,b);
}
//最小値更新
void chmin(long long &a,long long b){
  a=min(a,b);
}

//構造体
//Union-Find木
struct UnionFind{
  //メンバ変数 
  vector<long long>par; //par[i]:=頂点iの親
  vector<long long>s; //s[i]:=頂点iが属する集合の個数(iが根のとき)
  
  //コンストラクタ
  UnionFind(long long N):par(N),s(N){
    for(long long i=0;i<N;i++){
      par[i]=i; //最初はすべてが根として初期化
      s[i]=1;
    }
  }
  
  //メンバ関数
  //root(i):=頂点iの根
  long long root(long long i){
    if(par[i]==i) return i;
    par[i]=root(par[i]); //経路圧縮
    return root(par[i]); //再帰
  }
  //size[i]:=iが属する集合の個数
  long long size(long long i){
    return s[root(i)];
  }
  //same(x,y) xとyが同じ根を持つか(同じ集合に含まれるか)
  bool same(long long x, long long y){
    return root(x)==root(y);
  }
  //unite(x,y) xの根をyの根を繋げる(集合を合併)
  void unite(long long x, long long y){
    if(!same(x,y)){
      long long rx=root(x);
      long long ry=root(y);
      par[rx]=par[ry];
      s[ry]=s[rx]+s[ry];
    }
  }
};

//メイン関数
int main(){
  //doubleの桁数
  cout<<fixed<<setprecision(10);
  
  lin(N);lin(K);
  ll R=min(N,K);
  vlin(v,N);
  
  ll res=0;
  
  for(int A=0;A<=R;A++){
    for(int B=0;A+B<=R;B++){
      ll V=0; vl have(N,0);
      
      for(int i=0;i<A;i++){
        V+=v[i]; have[i]=v[i];
        //cout<<v[i]<<" ";
      }
      //cout<<":"<<" ";
      for(int j=0;j<B;j++){
        V+=v[N-1-j]; have[A+j]=v[N-1-j];
        //cout<<v[N-1-j]<<" ";
      }
      
      sort(all(have));
      
      rep(i,K-(A+B)){
        if(have[i]<0) V-=have[i];
      }
      
      chmax(res,V);
      //cout<<"::"<<V<<endl;
    }
  }
  
  cout<<res;
}