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: Wrong Answer)


import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.stream.IntStream;

public class Main {

	public static void main(String[] args) throws IOException {
		new Main().solve();
	}

	private void solve() throws IOException {
		try {
			//			solveA();
			//			solveB();
			//			solveC();
			solveD();
			//			 solveE();
			//			 solveF();
		} finally {
			if (in != null) {
				in.close();
			}
			if (out != null) {
				out.flush();
				out.close();
			}
		}

	}

	private void solveA() {

		out.println((nextInt() * 3 + nextInt()) / 2);
	}

	private void solveB() {
		int n = nextInt();
		String[][] s = new String[n][3];

		for (int i = 0; i < n; i++) {
			s[i][0] = next();
			s[i][1] = next();
			s[i][2] = Integer.toString(i + 1);
		}
		Arrays.sort(s,
				(x, y) -> {
					if (x[0].compareTo(y[0]) < 0) {
						return -1;
					} else if (x[0].compareTo(y[0]) > 0) {
						return 1;
					} else {
						return -Integer.compare(Integer.parseInt(x[1]), Integer.parseInt(y[1]));
					}
				});

		for (int i = 0; i < s.length; i++) {
			out.println(s[i][2]);
		}
	}

	private void solveC() {
		int n = nextInt();
		int m = nextInt();
		int[] k = new int[m];

		List<List<Integer>> s = new ArrayList<List<Integer>>();
		for (int i = 0; i < k.length; i++) {
			k[i] = nextInt();
			List<Integer> temp = new ArrayList<Integer>();
			for (int j = 0; j < k[i]; j++) {
				temp.add(nextInt());
			}
			s.add(temp);
		}
		int[] p = IntStream.range(0, m).map(i -> nextInt()).toArray();

		long res = 0;
		for (int i = 0; i < 1 << n; i++) {
			boolean isResult = true;
			for (int j = 0; j < m; j++) {
				List<Integer> sList = s.get(j);
				int compareBase = p[j];
				int compare = 0;
				for (Integer buttonNum : sList) {
					if ((i & (1 << (buttonNum - 1))) >= 1) {
						compare++;
					}
				}
				if (compareBase != compare % 2) {
					isResult = false;
					break;
				}
			}
			if (isResult) {
				res++;
			}

		}

		out.println(res);
	}

	private void solveD() {
		int n = nextInt();
		int k = nextInt();
		long[] wk = new long[n];
		for (int i = 0; i < n; i++) {
			wk[i] = nextInt();
		}
		long res = 0;
		//左からとれるだけ取る
		for (int l = 0; l <= k; l++) {
			//右からとれるだけ取る
			for (int r = 0; r <= k; r++) {
				//操作回数がkを超えるならbreak
				if (l + r > k) {
					break;
				}

				//現時点の手持ちの合計
				long now = 0;
				List<Long> s = new ArrayList<Long>();
				//左側から取得した分だけ足す
				for (int i = 0; i < Integer.min(l, n); i++) {
					long tmp = wk[i];
					now += tmp;
					s.add(tmp);
				}
				//右側から取得した分だけ足す
				for (int i = n - 1; i > (n - 1 - r >= 0 ? n - 1 - r : 0); i--) {
					long tmp = wk[i];
					now += tmp;
					s.add(tmp);
				}
				//余計なものを捨てるために、要らない順に並べる
				Collections.sort(s);

				//余分な取得物を捨てることができる回数
				int d = k - l - r;
				//この回数以上は捨てられないのでbreak
				for (int i = 0; i < Integer.min(d, s.size()); i++) {
					//
					//取り出した宝石の価値が負の値でないなら捨てる必要がない
					if (s.get(i) > 0) {
						break;
					}

					//負の価値の宝石は捨てる
					now -= s.get(i);
				}
				res = Long.max(res, now);
			}
		}
		out.println(res);
	}

	private void solveE() {
		int n = nextInt();

		out.println("");
	}

	private void solveF() {
		int n = nextInt();

		out.println("");
	}

	private final PrintWriter out = new PrintWriter(System.out);
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	private void skipUnprintable() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
	}

	public boolean hasNext() {
		skipUnprintable();
		return hasNextByte();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
}