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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author KharYusuf
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DEqueue solver = new DEqueue();
        solver.solve(1, in, out);
        out.close();
    }

    static class DEqueue {
        public void solve(int testNumber, FastReader s, PrintWriter w) {
            int n = s.nextInt(), k = s.nextInt(), a[] = new int[n], ans = 0;
            for (int i = 0; i < n; i++) a[i] = s.nextInt();
            for (int i = 0; i <= Math.min(n, k); i++) {
                for (int j = 0; i + j <= Math.min(n, k); j++) {
                    int curans = 0, remove = 0, num[] = new int[i + j], cur = 0;
                    for (int m = 0; m < i; m++) {
                        curans += a[m];
                        num[cur++] = a[m];
                    }
                    for (int m = 0; m < j; m++) {
                        curans += a[n - m - 1];
                        num[cur++] = a[n - m - 1];
                    }
                    func.sort(num);
                    for (int m = 0; m < cur && m < k - cur && num[m] < 0; m++) remove += num[m];
                    ans = Math.max(ans, curans - remove);
                }
            }
            w.println(ans);
        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private FastReader.SpaceCharFilter filter;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {

            if (numChars == -1)
                throw new InputMismatchException();

            if (curChar >= numChars) {

                curChar = 0;

                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }

                if (numChars <= 0)
                    return -1;
            }

            return buf[curChar++];
        }

        public int nextInt() {

            int c = read();

            while (isSpaceChar(c))
                c = read();

            int sgn = 1;

            if (c == '-') {
                sgn = -1;
                c = read();
            }

            int res = 0;

            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();

                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));

            return res * sgn;
        }

        public boolean isSpaceChar(int c) {

            if (filter != null)
                return filter.isSpaceChar(c);

            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);

        }

    }

    static class func {
        public static void sort(int arr[]) {
            int n = arr.length, mid, h, s, l, i, j, k;
            int[] res = new int[n];
            for (s = 1; s < n; s <<= 1) {
                for (l = 0; l < n - 1; l += (s << 1)) {
                    h = Math.min(l + (s << 1) - 1, n - 1);
                    mid = Math.min(l + s - 1, n - 1);
                    i = l;
                    j = mid + 1;
                    k = l;
                    while (i <= mid && j <= h) res[k++] = (arr[i] <= arr[j] ? arr[i++] : arr[j++]);
                    while (i <= mid) res[k++] = arr[i++];
                    while (j < n && j <= h) res[k++] = arr[j++];
                    for (k = l; k <= h; k++) arr[k] = res[k];
                }
            }
        }

    }
}