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: Wrong Answer)

use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

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

    if n % 2 == 0 {
        let mut dp = vec![0; n];
        dp[0] = a[0];
        dp[1] = a[1];
        for i in 2..n {
            dp[i] = std::cmp::max(dp[i - 1], dp[i - 2] + a[i]);
        }
        println!("{}", dp[n - 1]);
    } else {
        let none = -1_000_000_000_000;
        let mut dp = vec![vec![(none, none); n / 2 + 1]; n + 1];
        for i in 0..n / 2 + 1 {
            dp[i][0].1 = 0;
        }
        dp[0][1].0 = a[0];
        for i in 1..n {
            for j in 1..=std::cmp::min(n / 2, i) {
                dp[i][j].1 = std::cmp::max(dp[i - 1][j].0, dp[i - 1][j].1);
                dp[i][j].0 = dp[i - 1][j - 1].1 + a[i];
            }
        }
        // println!("{:?}", dp);
        println!("{}", std::cmp::max(dp[n - 1][n / 2].0, dp[n - 1][n / 2].1));
    }
}