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: Time Limit Exceeded)

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

dp = Array.new(n) {Array.new(n/2 + 2, -(10 ** 10))}
dp[0][0] = dp[1][0] = dp[-1][0] = dp[-2][0] = 0

if n.even?
    n.times do |i|
        dp[i][(i+1)/2] = [dp[i-1][(i+1)/2], dp[i-2][(i-1)/2] + a[i]].max if i > 0
        dp[i][i/2+1] = [dp[i-1][i/2+1], dp[i-2][i/2] + a[i]].max
    end
else
    n.times do |i|
        dp[i][i/2] = [dp[i-1][i/2], dp[i-2][i/2-1] + a[i]].max if i > 1
        dp[i][i/2+1] = [dp[i-1][i/2+1], dp[i-2][i/2] + a[i]].max
    end
end

puts dp[n-1][n/2]