Problem:
Score : 600 points

Problem Statement

Given is an integer sequence A_1, ..., A_N of length N.

We will choose exactly \left\lfloor \frac{N}{2} \right\rfloor elements from this sequence so that no two adjacent elements are chosen.

Find the maximum possible sum of the chosen elements.

Here \lfloor x \rfloor denotes the greatest integer not greater than x.

Constraints

2 \leq N \leq 2\times 10^5

|A_i|\leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the maximum possible sum of the chosen elements.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

12

Choosing 2, 4, and 6 makes the sum 12, which is the maximum possible value.

Sample Input 2

5
-1000 -100 -10 0 10

Sample Output 2

0

Choosing -10 and 10 makes the sum 0, which is the maximum possible value.

Sample Input 3

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

5000000000

Watch out for overflow.

Sample Input 4

27
18 -28 18 28 -45 90 -45 23 -53 60 28 -74 -71 35 -26 -62 49 -77 57 24 -70 -93 69 -99 59 57 -49

Sample Output 4

295

Solution in Julia (1.4.0) (judge verdict: Wrong Answer)


function main()
  n = parse(Int, readline())
  arr = parse.(Int, split(readline()))

  is_odd = n % 2 > 0

  space = is_odd ? 4 : 3
  dp = zeros(Int, n + space, 3)

  for (i, a) in enumerate(arr)
    i += space
    dp[i, 1] = dp[i - 2, 1] + a
    dp[i, 2] = max(dp[i - 2, 2], dp[i - 3, 1]) + a
    if is_odd
      dp[i, 3] = max(dp[i - 2, 3], dp[i - 3, 2], dp[i - 4, 1]) + a
    end
  end

  n += space
  println(
    is_odd ? max(dp[n - 2, 1], dp[n - 1, 2], dp[n, 3]) : max(dp[n, 2], dp[n - 1, 1])
  )
end

main()