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)

use proconio::*;
use std::collections::BTreeMap;

fn run() {
    input! {
        n: usize,
        mut a: [i64; n],
    }
    let mut memo = vec![];
    for _ in 0..2 {
        // 塗った数、直前に塗った、スコア
        let mut map = BTreeMap::<(usize, bool), i64>::new();
        map.insert((0, false), 0);
        let mut ans = vec![];
        ans.push(map.clone());
        for (i, &a) in a.iter().enumerate() {
            let mut next = BTreeMap::new();
            for ((c, p), s) in map {
                if c < i / 2 {
                    continue;
                }
                if !p {
                    let po = next.entry((c + 1, true)).or_insert(std::i64::MIN);
                    *po = std::cmp::max(*po, s + a);
                }
                let po = next.entry((c, false)).or_insert(std::i64::MIN);
                *po = std::cmp::max(*po, s);
            }
            map = next;
            ans.push(map.clone());
        }
        a.reverse();
        memo.push(ans);
    }
    let mut right = memo.pop().unwrap();
    right.reverse();
    let left = memo.pop().unwrap();
    let mut ans = std::i64::MIN;
    for (&(c, _), &x) in left[n].iter() {
        if c == n / 2 {
            ans = std::cmp::max(ans, x);
        }
    }
    for i in 1..=n {
        for (&(c, _), &x) in left[i - 1].iter() {
            for (&(d, _), &y) in right[n - i].iter() {
                if c + d == n / 2 {
                    ans = std::cmp::max(ans, x + y);
                }
            }
        }
    }
    println!("{}", ans);
}

fn main() {
    run();
}