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 Ruby (2.7.1) (judge verdict: Wrong Answer)

def max(a,b); a > b ? a : b; end
N = gets.to_i
A = gets.split.map(&:to_i)

dp = Array.new(4){ Array.new(N + 1, -Float::INFINITY) }
dp[0][0] = 0

ans = 0
(1 .. N).each do |i|
  a = A[i - 1]
  i_max = (i + 1) / 2
  i_min = N / 2 - (N - i + 1) / 2

  (i_min .. i_max).each do |t|
    (2 .. 4).each do |d|
      j = i - d
      next if j < 0
      
      j_max = (j + 1) / 2
      next if j_max - (t - 1) < 0
      
      dp[i_max - t][i] = max(dp[i_max - t][i], dp[j_max - (t - 1)][j] + a)
    end
    ans = max(ans, dp[i_max - t][i]) if t == N / 2 
  end
end
puts ans