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

if n.even?
    dp = Array.new(n, 0)
    n.times {|i| dp[i] = dp[i - 2] + a[i]}
    puts [dp[n - 1], dp[n - 2]].max
else
    dp = Array.new(3) {Array.new(n, 0)}
    (n/2).times {|i| dp[2][2*i] = dp[2][2*i-2] + a[2*i]}
    (n/2).times {|i| dp[1][2*i+1] = [dp[1][2*i-1], dp[2][2*i-2]].max + a[2*i+1]}
    (n/2).times {|i| dp[0][2*i+2] = [dp[0][2*i], dp[1][2*i-1], dp[2][2*i-2]].max + a[2*i+2]}
    puts [dp[0][n - 1], dp[1][n - 2], dp[2][n - 3]].max
end