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 Rust (1.42.0) (judge verdict: Accepted)

// Original: https://atcoder.jp/contests/abc162/submissions/11852555
use proconio::input;

fn main() {
    input! {
        n :usize,
        av: [i64; n],
    };

    let ans = if n % 2 == 0 {
        let mut dp = (0_i64, 0_i64);
        for (i, &a) in av.iter().enumerate() {
            if i % 2 == 0 {
                dp.0 = dp.0 + a;
            } else {
                dp.1 = std::cmp::max(dp.0, dp.1 + a);
            }
        }
        dp.1
    } else {
        let mut dp = (0_i64, 0_i64, 0_i64);
        for (i, &a) in av.iter().enumerate() {
            if i % 2 == 0 {
                dp.0 = dp.0 + a;
                if i != 0 {
                    dp.2 = std::cmp::max(dp.1, dp.2 + a);
                }
            } else {
                dp.1 = std::cmp::max(dp.0, dp.1 + a);
            }
        }
        dp.2
    };
    println!("{}", ans);
}