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: Accepted)

INF = - 10 ** 20
n = gets.to_i
nums = gets.split.map(&:to_i)
# [num][採用数ぴったりとの差(-1, 0, +1)][直前選んだかどうか]
dp = (n + 1).times.map { 3.times.map { Array.new(2, INF) } }
# 初期化
# dp[0 個目のnum][採用数はぴったり][直前は選んでいない]の最大値は 0
dp[0][1][0] = 0

# n個分ループする（i は遷移元の数字）
n.times do |i|
  # 最適化のため、事前に 1 行分だけ参照渡しをする
  dp_current = dp[i]
  dp_next = dp[i + 1]
  next_num = nums[i]
  3.times do |count|
    next_count = i.odd? ? count - 1 : count
    2.times do |selected|
      if selected == 1
        next if next_count < 0 || next_count > 2
        dp_next[next_count][0] = dp_current[count][1] if dp_next[next_count][0] < dp_current[count][1]
      else
        if next_count < 2
          dp_next[next_count + 1][1] = dp_current[count][0] + next_num if dp_next[next_count + 1][1] < dp_current[count][0] + next_num
        end
        if next_count > - 1
          dp_next[next_count][0] = dp_current[count][0] if dp_next[next_count][0] < dp_current[count][0]
        end
      end
    end
  end
end

puts dp.last[1].max