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: Time Limit Exceeded)

use proconio::input;

fn solve(k: usize, sum: i64, v: &Vec<i64>) -> i64 {
    if k == 0 {
        return sum;
    }

    if k > v.len() / 2 + 1 {
        return std::i64::MIN;
    }

    if v.len() == 0 {
        return std::i64::MIN;
    }

    if v.len() == 1 {
        if k == 1 {
            return sum + v[0];
        }
        return std::i64::MIN;
    }

    std::cmp::max(
        solve(k - 1, sum + v[0], &v[2..].to_vec()),
        solve(k, sum, &v[1..].to_vec()),
    )
}

fn main() {
    input! {
        n: usize,
        a_vec: [i64; n],
    }

    println!("{}", solve(n / 2, 0, &a_vec));
}