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: Runtime Error)

#[allow(dead_code)]
fn is_whitespace(c: &char) -> bool {
    c.is_whitespace()
}

#[allow(dead_code)]
fn is_eol(c: &char) -> bool {
    ['\r', '\n'].contains(c)
}

#[allow(unused_macros)]
macro_rules! input {
    ($var:ident, $($r:tt)*) => {
        input_inner!{$var, $($r)*}
    };

    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        #[allow(unused_mut, unused_variables, bare_trait_objects)]
        let mut next = move |f: &Fn(&char)->bool| -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|f(&c))
                .take_while(|c|!f(&c))
                .collect()
        };
        input_inner!{next, $($r)*};
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; eol ]) => {
        $next(&is_eol).split(' ').map(|x| x.parse::<$t>().unwrap()).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, parser) => {
        $next
    };

    ($next:expr, $t:ty) => {
        $next(&is_whitespace).parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BinaryHeap, HashMap, HashSet};

#[allow(dead_code)]
struct Edge {
    from: usize,
    to: usize,
}

#[allow(dead_code)]
type Graph = Vec<Vec<usize>>;
#[allow(dead_code)]
const M: usize = 1_000_000_007;

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

    let mut dp = vec![vec![None; 3]; n];
    dp[0][0] = Some(a[0]);
    dp[1][1] = Some(a[1]);
    dp[2][2] = Some(a[2]);

    for i in 0..n {
        for j in 0..3 {
            if dp[i][j].is_none() {
                continue;
            }
            for k in 0..3 {
                if i + 2 + k >= n {
                    continue;
                }
                if j + k > 2 {
                    continue;
                }
                let next = dp[i][j].unwrap() + a[i + 2 + k];
                if let Some(x) = dp[i + 2 + k][j + k] {
                    dp[i + 2 + k][j + k] = Some(max(next, x));
                } else {
                    dp[i + 2 + k][j + k] = Some(next);
                }
            }
        }
    }
    if n % 2 == 0 {
        println!("{}", max(dp[n - 2][0].unwrap(), dp[n - 1][1].unwrap()));
    } else {
        let mut ans = max(dp[n - 3][0].unwrap(), dp[n - 2][1].unwrap());
        ans = max(ans, dp[n - 1][2].unwrap());
        println!("{}", ans);
    }
}