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 Lua (5.3.2) (judge verdict: Accepted)

local DBG = false
local function dbgpr(...)
    if DBG then
        io.write("[dbg]")
        print(...)
    end
end

local function dbgpr_t(tbl, use_pairs)
    if DBG then
        local enum = ipairs
        if use_pairs then
            enum = pairs
        end
        dbgpr(tbl)
        io.write("[dbg]")
        for i,v in enum(tbl) do
            io.write(i)
            io.write(":")
            io.write(tostring(v))
            io.write(" ")
        end
        print("")
    end
end

----
local N, Kmax = io.read("n", "n")
local V = {}
for i=1,N do
    V[i] = io.read("n")
end

local lim_left = math.min(N, Kmax)
local ans = 0
for a=0, lim_left do
    local got_left = {}
    local lim_right = math.min(N-a, Kmax-a)
    assert(lim_right >= 0)
    for i=1,a do
        got_left[i] = V[i]
    end
    for b=0, lim_right do
        local got = {table.unpack(got_left)}
        for i=1,b do
            got[#got + 1] = V[N+1-i]
        end
        table.sort(got)
        local ndiscard = Kmax - a - b
        local ndiscard_orig = ndiscard
        assert(ndiscard >= 0)
        local value = 0
        for i=1, #got do
            if got[i] < 0 and ndiscard > 0 then
                ndiscard = ndiscard - 1
            else
                value = value + got[i]
            end
        end
        dbgpr(a, b, #got, ndiscard_orig, value)
        ans = math.max(ans, value)
    end
end
print(ans)