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)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)

#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)x.size()
#define beg(x) x.begin()
#define en(x) x.end()
#define all(x) beg(x), en(x)
#define resz resize

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
const ld PI = 4*atan((ld)1);

template<class T> void ckmin(T &a, T b) { a = min(a, b); }
template<class T> void ckmax(T &a, T b) { a = max(a, b); }

namespace input {
    template<class T> void re(complex<T>& x);
    template<class T1, class T2> void re(pair<T1,T2>& p);
    template<class T> void re(vector<T>& a);
    template<class T, size_t SZ> void re(array<T,SZ>& a);

    template<class T> void re(T& x) { cin >> x; }
    void re(double& x) { string t; re(t); x = stod(t); }
    void re(ld& x) { string t; re(t); x = stold(t); }
    template<class Arg, class... Args> void re(Arg& first, Args&... rest) {
        re(first); re(rest...);
    }

    template<class T> void re(complex<T>& x) { T a,b; re(a,b); x = cd(a,b); }
    template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
    template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
    template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
}

using namespace input;

namespace output {
    template<class T1, class T2> void pr(const pair<T1,T2>& x);
    template<class T, size_t SZ> void pr(const array<T,SZ>& x);
    template<class T> void pr(const vector<T>& x);
    template<class T> void pr(const set<T>& x);
    template<class T1, class T2> void pr(const map<T1,T2>& x);

    template<class T> void pr(const T& x) { cout << x; }
    template<class Arg, class... Args> void pr(const Arg& first, const Args&... rest) {
        pr(first); pr(rest...);
    }

    template<class T1, class T2> void pr(const pair<T1,T2>& x) {
        pr("{",x.f,", ",x.s,"}");
    }
    template<class T> void prContain(const T& x) {
        pr("{");
        bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0; // const needed for vector<bool>
        pr("}");
    }
    template<class T, size_t SZ> void pr(const array<T,SZ>& x) { prContain(x); }
    template<class T> void pr(const vector<T>& x) { prContain(x); }
    template<class T> void pr(const set<T>& x) { prContain(x); }
    template<class T1, class T2> void pr(const map<T1,T2>& x) { prContain(x); }

    void ps() { pr("\n"); }
    template<class Arg> void ps(const Arg& first) {
        pr(first); ps(); // no space at end of line
    }
    template<class Arg, class... Args> void ps(const Arg& first, const Args&... rest) {
        pr(first," "); ps(rest...); // print w/ spaces
    }
}

using namespace output;

namespace io {
    void setIn(string s) { freopen(s.c_str(),"r",stdin); }
    void setOut(string s) { freopen(s.c_str(),"w",stdout); }
    void setIO(string s = "") {
        ios_base::sync_with_stdio(0); cin.tie(0); // fast I/O
        if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
    }
}

using namespace io;

template<class T> T invGeneral(T a, T b) {
    a %= b; if (a == 0) return b == 1 ? 0 : -1;
    T x = invGeneral(b,a);
    return x == -1 ? -1 : ((1-(ll)b*x)/a+b)%b;
}

template<class T> struct modular {
    T val;
    explicit operator T() const { return val; }
    modular() { val = 0; }
    template<class U> modular(const U& v) {
        val = (-MOD <= v && v <= MOD) ? v : v % MOD;
        if (val < 0) val += MOD;
    }
    friend ostream& operator<<(ostream& os, const modular& a) { return os << a.val; }
    friend bool operator==(const modular& a, const modular& b) { return a.val == b.val; }
    friend bool operator!=(const modular& a, const modular& b) { return !(a == b); }

    modular operator-() const { return modular(-val); }
    modular& operator+=(const modular& m) { if ((val += m.val) >= MOD) val -= MOD; return *this; }
    modular& operator-=(const modular& m) { if ((val -= m.val) < 0) val += MOD; return *this; }
    modular& operator*=(const modular& m) { val = (ll)val*m.val%MOD; return *this; }
    friend modular exp(modular a, ll p) {
        modular ans = 1; for (; p; p /= 2, a *= a) if (p&1) ans *= a;
        return ans;
    }
    friend modular inv(const modular& a) { return invGeneral(a.val,MOD); }
    // inv is equivalent to return exp(b,b.mod-2) if prime
    modular& operator/=(const modular& m) { return (*this) *= inv(m); }

    friend modular operator+(modular a, const modular& b) { return a += b; }
    friend modular operator-(modular a, const modular& b) { return a -= b; }
    friend modular operator*(modular a, const modular& b) { return a *= b; }

    friend modular operator/(modular a, const modular& b) { return a /= b; }
};

typedef modular<int> mi;
typedef pair<mi,mi> pmi;
typedef vector<mi> vmi;
typedef vector<pmi> vpmi;
const int N = 1e2 + 10;
int n, k, a[N], b[N], ans;
int main()
{
    re(n,k);
    for (int i = 0; i < n; ++i) re(a[i]);
    for (int rock = 1; rock <= min(n,k); ++rock) {
        for (int l = 0; l <= rock; ++l) {
            int res = 0, ct = 0, tmp = k - rock;
            for (int i = 0; i < l; ++i) b[ct++] = a[i];
            for (int i = n-1; i >= n - rock + l; --i) b[ct++] = a[i];
            sort(b, b + ct);
            for (int i = 0; i < min(tmp, ct); ++i) if (b[i] < 0) b[i] = 0; else break;
            for (int i = 0; i < ct; ++i) res += b[i];
            ans = max(ans, res);
        }
    }
    ps(ans);
    return 0;
}