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)

n = gets.to_i
nums = gets.split.map(&:to_i)
length = nums.length
# dp[i 個目の num まで][前の数字を選んだか][採用回数]の出現回数
dp = (length + 1).times.map { 2.times.map { Array.new(length + 1, -1) } }
# 初期化
# dp[0 個目の num まで][前の数字を選んだか][採用回数]の最大値は 0
dp[0][0][0] = 0

# 桁数分ループする（i は遷移元の桁）
length.times do |i|
  dp_current = dp[i]
  # 最適化のため、事前に 1 行分だけ参照渡しをする
  # dp_next と dp[i + 1] の object_id は同じ
  dp_next = dp[i + 1]
  2.times do |selected|
    limit = (i / 2 + 1)
    range = ([limit - 2, 0].max..limit)
    range.each do |count|
      current_sum = dp_current[selected][count]
      next if current_sum == -1
      2.times do |next_selected|
        next if selected == 1 && next_selected == 1
        current_sum += nums[i] if next_selected == 1
        count += 1 if next_selected == 1
        dp_next[next_selected][count] = current_sum if dp_next[next_selected][count] < current_sum
      end
    end
  end
end

puts [dp[n][0][n / 2], dp[n][1][n / 2]].max