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)

use proconio::{input, fastout};
use std::cmp::max;


#[fastout]
fn main() {
    input! {
        n: usize,
        aaa: [isize; n],
    }

    if n % 2 == 0 {
        let mut fwd = vec![0];
        let mut bwd = vec![0];
        let acc_func = |t: &mut isize, &a| {
            *t += a;
            Some(*t)
        };
        fwd.extend(aaa.iter().step_by(2).scan(0, acc_func));
        bwd.extend(aaa.iter().rev().step_by(2).scan(0, acc_func));
        let ans = fwd.iter().zip(bwd.iter().rev()).map(|(f, b)| f + b).max().unwrap();
        println!("{:?}", ans);
    } else {
        const NINF: isize = -1_000_000_000_000_000_000;
        //                                  [直前は取ってない]  [直前に取った]
        // 連続して取らないことができる残り回数  0, 1, 2           0, 1, 2
        let mut dp = (NINF, NINF, 0isize, NINF, NINF, NINF);
        for a in aaa {
            dp = (
                max(dp.3, dp.1),
                max(dp.4, dp.2),
                dp.5,
                dp.0 + a,
                dp.1 + a,
                dp.2 + a,
            );
        }
        println!("{:?}", max(dp.1, dp.3));
    }
}