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 std::io::stdin;
use std::str::FromStr;
use std::cmp::max;

fn read_line<T>() -> Vec<T>
where
    T: FromStr,
    <T as std::str::FromStr>::Err: std::fmt::Debug,
{
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    s.trim()
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect()
}

fn main() {
    // DP $B;H$&LdBj$C$]$$(B
    let in_vec: Vec<i64> = read_line();
    let n = in_vec[0];
    let a: Vec<i64> = read_line();

    // $B2?2s(B +2 $B$G$-$k$+(B
    let plus_2_num = 2 + (n % 2);

    let mut dp: Vec<Vec<Option<i64>>> = vec![vec![None;plus_2_num as usize];(n/2) as usize];

    for i in 0..n/2 {
        let i = i as usize;
        for j in 0..plus_2_num {
            let j = j as usize;
            if i == 0 {
                dp[i][j] = Some(a[2*i+j]);
            } else {
                let better_num;
                if j > 0 {
                    better_num = Some(max(dp[i-1][j].unwrap(), dp[i-1][j-1].unwrap()));
                } else {
                    better_num = dp[i-1][j];
                }
                dp[i][j] = Some(better_num.unwrap() + a[2*i+j]);
            }
        }
    }

    let mut res = dp[(n/2-1) as usize][0].unwrap();
    for i in 1..plus_2_num {
        res = max(res, dp[(n/2-1) as usize][i as usize].unwrap());
    }

    println!("{}", res);
}