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 <iomanip>      // std::setprecision
using namespace std;
#define what_is(x) cerr << #x << " is " << x << endl;
#define bit(x,i) (x&(1<<i))  //select the bit of position i of x
#define lowbit(x) ((x)&((x)^((x)-1))) //get the lowest bit of x
#define hBit(msb,n) asm("bsrl %1,%0" : "=r"(msb) : "r"(n)) //get the highest bit of x, maybe the fastest
#define max(a,b) (a<b?b:a)
#define IN(i,l,r) (l<i&&i<r) //the next for are for checking bound
#define LINR(i,l,r) (l<=i&&i<=r)
#define LIN(i,l,r) (l<=i&&i<r)
#define INR(i,l,r) (l<i&&i<=r)
#define F(i,L,R) for (int i = L; i < R; i++) //next four are for "for loops"
#define FE(i,L,R) for (int i = L; i <= R; i++)
#define FF(i,L,R) for (int i = L; i > R; i--)
#define FFE(i,L,R) for (int i = L; i >= R; i--)
#define char2Int(c) (c-'0')
#define lastEle(vec) vec[vec.size()-1] 
#define SZ(x) ((int)((x).size()))
#define REMAX(a,b) (a)=max((a),(b)) // set a to the maximum of a and b
#define REMIN(a,b) (a)=min((a),(b));
#define FENWICK(k) (k&-k)
#define FOREACH(i,t) for (auto i=t.begin(); i!=t.end(); i++) // traverse an STL data structure
#define ALL(c) (c).begin(),(c).end() //handy for function like "sort()"
#define PRESENT(c,x) ((c).find(x) != (c).end()) 
#define CPRESENT(c,x) (find(ALL(c),x) != (c).end()) 
#define LL long long //data types used often, but you don't want to type them time by time
#define ull unsigned long long
#define ui unsigned int
#define us unsigned short
#define IOS ios_base::sync_with_stdio(0); //to synchronize the input of cin and scanf
#define INF 1001001001
#define PI 3.1415926535897932384626
//for map, pair
#define mp make_pair
#define fi first
#define se second
//for vectors
#define pb push_back
typedef int elem_t;
typedef vector<int> vi; 
typedef vector<vi> vvi; 
typedef pair<int,int> ii; 
// directions
const int fx[4][2] = {{0,1}, {0,-1}, {1,0}, {-1,0}};
const int fxx[8][2] = {{0,1}, {0,-1}, {1,0}, {-1,0}, {1,1}, {1,-1}, {-1,1}, {-1,-1}};
const int rogi[4][2] = { {1,1}, {1,-1}, {-1,1}, {-1,-1} };

int n,k;
LL val[111];
LL best = 0;

vector<LL> vals;
void dane() {
	cin>>n>>k;
	F(i,0,n) cin>>val[i];

	F(ile_lewa,0,n) {
		F(ile_prawa,0,(n-ile_lewa-1)) {
			LL sum = 0;
			if(ile_lewa + ile_prawa > k || ile_lewa + ile_prawa > n) {
				continue;
			}
			//cout<<"Ile lewa: "<<ile_lewa<<" "<<ile_prawa<<endl;
			vals.clear();
			//dodajemy elementy
			F(i,0,ile_lewa) {
				vals.push_back(val[i]);
				sum+=val[i];
			}
			F(i,0,ile_prawa) {
				vals.push_back(val[n-i-1]);
				sum+=val[n-i-1];
			}
			best = max(best, sum);
			//cout<<"SUM : "<<sum<<endl;
			sort(ALL(vals));
			int kroki = min((int)vals.size(),k- ile_lewa - ile_prawa);
			//poki mozna tu usuwamy najmniejszy
			F(i,0,kroki) {
				sum-=vals[i];
				best = max(best, sum);
			}
		}
	}
	cout<<best<<endl;

}    


int main() {
	IOS
	// int t;
	// cin>>t;
	// F(i,0,t)
	    dane();
    return 0;
}