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::collections::HashMap;

const INF : i64 = 1i64<<61;

fn solve(n : usize, k : usize, a : &Vec<i64>, memo : &mut HashMap<(usize, usize), i64>) -> i64 {
    if k == 0 { return 0 }
    else if n/2 > k || (n+2)/2 < k { return -INF }
    else {
        match memo.get(&(n, k)) {
            Some(ans) => return *ans,
            None =>  {
                let v1 = a[n] + solve(n-2, k-1, a, memo);
                let v2 = solve(n-1, k, a, memo);
                let res = std::cmp::max(v1, v2);
                memo.insert((n,k), res);
                return res
            }
        }
    }
}

fn main() {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    use std::iter::FromIterator;
    let mut iter = input.split_whitespace();
    let n : usize = iter.next().unwrap().parse().unwrap();
    let a : Vec<i64> = Vec::from_iter(iter.map(|s| s.parse().unwrap()));

    let mut memo = HashMap::new();
    println!("{}", solve(n-1, n/2, &a, &mut memo))
}