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 <iostream>
#include <iomanip>
#include <fstream>
#include <string>
#include <array>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <stack>
#include <queue>
#include <bitset>
#include <tuple>
#include <cmath>
#include <complex>
#include <algorithm>
#include <utility>
#include <regex>
#include <cstdint>
#include <numeric>
#include <functional>
#include <cassert>

using namespace std;
using ll  = long long;
using ull = unsigned long long;

namespace utils{
#define ALL(x) begin(x), end(x)
#define RALL(x) rbegin(x), rend(x)
    template<class T, class Compare> using p_queue = priority_queue<T, vector<T>, Compare>;
    template<class T> using min_queue = p_queue<T, greater<T>>;
    template<class T> using max_queue = p_queue<T, less<T>>;

    template<class T> inline bool UPMIN(T& X, const T& A){ if(X > A) {X = A; return true;} return false; }
    template<class T> inline bool UPMAX(T& X, const T& A){ if(X < A) {X = A; return true;} return false; }

    template<class T>
    vector<T> VEC(int n, T t){
        return vector<T>(n, t);
    }
    template<class ...Ts>
    auto VEC(int n, Ts ... ts){
        return vector<decltype(VEC(ts...))>(n, VEC(ts...));
    }

    template <class S, class T>
    istream& operator>>(istream& is, tuple<S, T>& t){
        return is >> get<0>(t) >> get<1>(t);
    }

    template <class S, class T, class U>
    istream& operator>>(istream& is, tuple<S, T, U>& t){
        return is >> get<0>(t) >> get<1>(t) >> get<2>(t);
    }

    template <class S, class T>
    ostream& operator<<(ostream& os, const tuple<S, T>& t){
        return os << get<0>(t) << ' ' <<  get<1>(t);
    }

    template <class S, class T, class U>
    ostream& operator<<(ostream& os, const tuple<S, T, U>& t){
        return os << get<0>(t) << ' ' <<  get<1>(t) << ' ' <<  get<2>(t);
    }

    template<class T>
    istream& operator>>(istream& is, vector<T>& v){
        for (auto &&x : v) { is >> x; } return is;
    }

    template<class T>
    ostream& operator<<(ostream& os, const vector<T>& v){
        auto p = v.begin();
        assert(p != v.end());
        os << *p++;
        while(p != v.end()){
            os << ' ' << *p++;
        }
        return os ;
    }

    template<class T>
    ostream& operator<<(ostream& os, const vector<vector<T>>& v){
        auto p = v.begin();
        assert(p != v.end());
        os << *p++;
        while(p != v.end()){
            os << '\n' << *p++;
        }
        return os;
    }

    template<class ...Ts>
    ostream& operator<<(ostream& os, const vector<tuple<Ts...>>& v){
        auto p = v.begin();
        assert(p != v.end());
        os << *p++;
        while(p != v.end()){
            os << '\n' << *p++;
        }
        return os;
    }

    constexpr long long INFLL   = 1'000'000'000'000'000'000ll;
    constexpr int       INF     = 1'000'000'000;
    constexpr double    PI      = 3.14'159'265'358'973;
    constexpr double    EPS     = 1e-10;
}
using namespace utils;


class solver{
    istream& is;
    ostream& os;

public:
    solver(istream& I, ostream& O) :is(I), os(O) {}
    ll N, K;
    vector<ll> V;

    bool input() {
        N = K = 0;
        is >> N >> K;
        V.resize(N);
        is >> V;
        return !!is;
    }

    void run();
};

int main(int argc, char *argv[]) {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << setprecision(16) << scientific;
#ifdef XELMH
    string test_cases = "test_D.txt";
    cerr << test_cases << " -->" << endl;
    auto fs = fstream(test_cases, fstream::in);
    int loop = 0;
    while(fs) {
        loop++;
        cout << '#' << loop << "#------\n";
        solver(fs, cout).run();
    }
    if(loop <= 1) {
        cout << "===" << endl;
        while(cin) solver(cin, cout).run();
    }
#else
    solver(cin, cout).run();
#endif
    return 0;
}


void solver::run(){
    if(!input()) return;
    ll ans = 0;
    for (int l = 0; l <= N; ++l) {
        for (int r = 0; r <= K - l; ++r) {
            if(r >= N) break;
            if(l + r > N) break;
            min_queue<ll> get;
            for (int i = 0; i < l; ++i) { get.push(V[i]); } // end i
            for (int i = 0; i < r; ++i) { get.push(V[N-i-1]); } // end i
            ll remK = K - l - r;
            while(remK > 0 and !get.empty()){
                if(get.top() > 0) break;
                get.pop(); remK--;
            }
            ll sum = 0;
            while(!get.empty()){
                sum += get.top(); get.pop();
            }
            UPMAX(ans, sum);
        } // end r
    } // end l;
    os << ans << endl;

}