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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let n = read::<usize>();
    let a = read_vec::<i64>();
    if n == 2 {
        println!("{}", max(a[0], a[1]));
        return;
    }
    let mut dp = vec![vec![-std::i64::MAX; 3]; n];
    dp[0][0] = a[0];
    dp[1][1] = a[1];
    dp[2][2] = a[2];
    for i in 2..n {
        for j in 0..3 {
            if dp[i - 2][j] != -std::i64::MAX {
                dp[i][j] = max(dp[i - 2][j] + a[i], dp[i][j]);
            }
            if j < 2 && i >= 3 && dp[i - 3][j] != -std::i64::MAX {
                dp[i][j + 1] = max(dp[i - 3][j] + a[i], dp[i][j + 1]);
            }
            if j < 1 && i >= 4 && dp[i - 4][j] != -std::i64::MAX {
                dp[i][j + 2] = max(dp[i - 4][j] + a[i], dp[i][j + 2]);
            }
        }
    }
    if n % 2 == 0 {
        println!("{}", max(dp[n - 1][1], dp[n - 2][1]));
    } else {
        println!("{}", max(max(dp[n - 1][2], dp[n - 2][1]), dp[n - 3][0]));
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}