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: Wrong Answer)

#include <bits/stdc++.h>
#include <algorithm>
#include <cmath>
#include <set>
#include <cstdio>
#include <vector>
#include <iostream>
#include <utility>
#include <queue>
#include <map>

#define fir first
#define sec second
#define sz(s) (s).size()
#define pb push_back
#define get(n) scanf("%d",&n);
#define gets(s) string s;cin >> (s);
#define prfi(n) printf("%d", &n);
#define prfd(n) printf("%lf", &n);
#define All(s) (s).begin(), (s).end()
#define rep(i,j) for(int (i)=0;(i)<(j);(i)++)
#define For(i,j,k) for(int (i)=(j);(i)<(k);(i)++)
#define drep(i,j) for(int (i)=(j);(i)>=0;(i)--)
#define Ford(i,j,k) for(int (i)=(j);i>=(k);i--)
#define vfor(c,v) for(auto (c): (v))
#define lpi(n) for(int i=0;i<(n);i++)
#define lpj(n) for(int j=0;j<(n);j++)
#define lpz(n) for(int z=0;z<(n);z++)
#define mem(a,b) memset(a,b,sizeof(a));
#define dump(x)  std::cout << #x << " = " << (x) << std::endl;
#define debug(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using ull = unsigned long long int;
using ll = long long;
using ld = long double;
using pii = std::pair<int,int>;
using pll = std::pair<ll, ll>;
using vi = std::vector<int> ;
using vvi = std::vector<vi> ;
using vll = std::vector<ll>;
using vvll = std::vector<vll>;
using vd = std::vector<double> ;
using vvd = std::vector<vd> ;
using qi = std::queue<int> ;
using vpii = std::vector<std::pair<int, int> >;
using vpll = std::vector<pll>;
using namespace std;

const int Mod = (1e9) + 7;
const int INF = 1e9 + 19;
const ll INFL = 1e18 + 19;
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
const int dx2[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1};
const int dy2[] = {1, 0, -1, 1, 0, -1, 1, 0, -1};
//_____________________________________Templates_________________________________________//

template<class T1, class T2> inline void chmin(T1 &a, T2 b){if(a > b) a = b;}
template<class T1, class T2> inline void chmax(T1 &a, T2 b){if(a < b) a = b;}
template<class T> inline void pri(T a){cout << a << endl;}
template<class Z> using vec = vector<Z>;
//mainly use for dynamic prog
template<class T1, class T2>
void update(T1 &a, T2 b){
  a += b;
  if(a > Mod) a %= Mod;
}

inline void IN(void){
  return;
}

template <typename First, typename... Rest>
void IN(First& first, Rest&... rest){
  cin >> first;
  IN(rest...);
  return;
}

inline void OUT(void){
  cout << "\n";
  return;
}

template <typename First, typename... Rest>
void OUT(First first, Rest... rest){
  cout << first << " ";
  OUT(rest...);
  return;
}

bool pairsort(pll pl, pll pr){
  if(pl.first == pr.first)return pl.second > pr.second;
  return pl.first < pr.first;
}

int cntbit(ll a,int n,int j){int res = 0;For(i,j,n){if(a>>i & 1){res++;}}return res;}
vector<int> make_bit(int a){vector<int> res; lpi(32)if(a&(1<<i))res.pb(i);return res;}
bool stdbit(int a, int b){return a&(1 << b); }
int GCD(int a, int b){if(b > a)return GCD(b,a);if(a%b == 0)return b;else return GCD(b, a%b);}
int LCM(int a, int b){return a*b/GCD(a,b);}
int roundup(int a, int b){if(a % b == 0)return a/b;else return (a+b)/b;}
int rounddown(int a, int b){if(a%b == 0)return a/b;else {return (a-b)/b;}}
ll pow(ll a, ll n){ll res = 1;while(n > 0){if(n&1)res *= a; a *= a; n = n >> 1;}return res;}
ll GetDiviserCount(ll N)//約数の個数
{
  ll res = 1;
  For(i,2,sqrt(N)+1)
  {
    ll cnt = 0;
    while(N%i == 0)
    {
      cnt++;
      N /= i;
    }
    res *= (cnt + 1);
    if(N == 1)break;
  }
  if(N != 1)res *= 2;
  return res;
}
vll GetDivisor(ll N)//約数列挙
{
  vll res;
  for(ll i = 1;i*i <= N;i++)
  {
    if(N%i == 0)
    {
      res.pb(i);
      if(i*i != N)res.pb(N/i);
    }
  }
  sort(All(res));
  return res;
}
//_____________________　following sorce code_________________________//
const int max_n = 3 * (1e5) + 1;
const int max_m = 83 * (1e5) + 1;

int n,m,k;
ll N;
int h,w;
vvi tree;
string S;
int a,b,c;
vi v;
int ans;

int main () {
  cin.tie(0);
  ios::sync_with_stdio(false);
  IN(n,k);
  vll v(n);
  rep(i,n)IN(v[i]);
  ll ans = 0;
  For(i,1,k+1)
  {
    rep(j,i+1)
    {
      ll tmp = 0;
      vi u;
      int d = k - i;
      rep(l,j)
      {
        tmp += v[l];
        //cout << "v[" << l << "] = " << v[l] << endl;
        u.pb(v[l]);
      }
      rep(l,i-j)
      {
        tmp += v[n-1-l];
        //cout << "v[" << l  << "] = " << v[l] << endl;
        u.pb(v[n-1-l]);
      }
      sort(All(u));
      rep(l,min((int)(sz(u)), d))
      {
        if(u[l] > 0)break;
        tmp -= u[l];
      }
      //dump(j);
      //dump(i-j);
      //dump(tmp);
      chmax(ans, tmp);
    }
  }
  
  cout << ans << endl;
  //for(auto c : ans){cout << c << endl;}
  //cout << fixed << setprecision(15) << ans << endl;
  return 0;
}