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**15
n = gets.to_i
a = gets.split.map(&:to_i)

if n % 2 == 0
  evn = Array.new(n+1,0)
  odd = Array.new(n+1,0)
  for i in 0..(n-1)
    evp = odp = 0
    (i % 2 == 0) ? (evp = a[i]):(odp = a[i])
    evn[i+1] = evn[i] + evp
    odd[i+1] = odd[i] + odp
  end
  ans = -INF
  for i in 0..(n-1)
    if i % 2 == 0
      ans = [ans, evn[i]+odd[n]-odd[i]].max
    end
  end
  ans = [ans, evn[i]].max
  puts ans
else
  dp = Array.new(n).map{Array.new(3).map{Array.new(2,-INF)}}
  ans = -INF
  for i in 0..2
    dp[i][i][0] = a[i]
    dp[i][i][1] = 1
  end
  for i in 0..(n-1)
    for j in 0..2
      if i > j && (i % 2 == j % 2) && dp[i-2][j][1] < n/2
        for k in 0..j
          dp[i][j][0] = [dp[i][j][0], dp[i-2-k][j-k][0]+a[i]].max
        end
        dp[i][j][1] = dp[i-2][j][1] + 1
      end
      ans = [ans, dp[i][j][0]].max if dp[i][j][1] == n/2
    end
  end
  puts ans
end