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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
import itertools
sys.setrecursionlimit(10**5)
stdin = sys.stdin
bisect_left = bisect.bisect_left
bisect_right = bisect.bisect_right
def LI(): return list(map(int, stdin.readline().split()))
def LF(): return list(map(float, stdin.readline().split()))
def LI_(): return list(map(lambda x: int(x)-1, stdin.readline().split()))
def II(): return int(stdin.readline())
def IF(): return float(stdin.readline())
def LS(): return list(map(list, stdin.readline().split()))
def S(): return list(stdin.readline().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')

#A
def A():
    a, p = LI()
    print((a*3+p)//2)
    return

#B
def B():
    n = II()
    sp = []
    for i in range(n):
        sp.append(input().split())
    another = [(s[0], -1 * int(s[1]),num) for num,s in enumerate(sp)]
    another.sort()
    for s,p,q in another:
        print(q+1)

    return

#C
def C():
    n, m = LI()
    ks = LIR_(m)
    p = LI()
    patern = itertools.product(range(2), repeat=n)
    ans = 0
    for on in patern:
        ki = [0 for i in range(m)]
        for num, oni in enumerate(on):
            if oni:
                for i,ksi in enumerate(ks):
                    ksi = ksi[1::]
                    if num in ksi:
                        ki[i] ^= 1
        for num, pi in enumerate(p):
            if pi != ki[num]:
                break
        else:
            ans += 1
    print(ans)

            
    return

#D
def D():
    N, K = LI()
    v = LI()
    ans = 0
    for k in range(K // 2 + 1):
        #print(k)
        for n in range(1, N - 1):
            v0 = v[:n]
            v1 = v[n:]
            v1 = v1[::-1]
            for v0i in range(n + 1):
                for v1i in range(N - n + 1):
                    if v0i + v1i > K :
                        break
                    if v0i == v1i == 0:
                        continue
                    v0l = v0[:v0i]
                    v1l = v1[:v1i]
                    v0l.sort()
                    v1l.sort()
                    sumvl = sum(v0l) + sum(v1l)
                    for v0k in range(k + 1):
                        for v1k in range(k - v0k + 1):
                            if v0i + v0k + v1i + v1k > K:
                                break
                            ans = max(ans, sumvl - sum(v1l[:v1k]) - sum(v0l[:v0k]))
                            #print(v0,v1,v0i,v1i,v0l,v1l,v0k,v1k,sumvl - sum(v1l[:v1k]) - sum(v0l[:v0k]))
    print(ans)
                

    return

#E
def E():
    return

#F
def F():
    return

#Solve
if __name__ == '__main__':
    D()