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)

macro_rules! read_line {
    () => {{
        let mut input = String::new();
        std::io::stdin().read_line(&mut input).unwrap();
        input
    }};
}
pub fn read_and_parse<T: std::str::FromStr>() -> T {
    read_line!().trim().parse().ok().unwrap()
}
pub fn read_to_vec<T: std::str::FromStr>() -> Vec<T> {
    read_line!().split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
use std::cmp::max;
fn main() {
    let n: usize = read_and_parse();
    let a: Vec<i64> = read_to_vec();

    let mut dp = Vec::with_capacity(n+2);
    dp.resize(n+2, [0, 0]);

    for i in 1..=n {
        if i % 2 == 1 {
            dp[i + 1] = [max(dp[i - 1][0] + a[i - 1], dp[i][1]), dp[i - 1][1] + a[i - 1]];
        } else {
            dp[i + 1] = [max(dp[i - 1][0] + a[i - 1], dp[i][0]),
                max(dp[i - 1][0] + a[i - 1], dp[i][1])];
        }
    }

    println!("{}", dp.last().unwrap()[(n + 1) % 2]);
}