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::io::{self, BufRead};
use std::cmp::max;

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let line = lines.next().unwrap().unwrap();
  let n: usize = line.parse().unwrap();
  let line = lines.next().unwrap().unwrap();
  let xs: Vec<i64> = line.split(' ').map(|x| x.parse().unwrap()).collect();
  let mut memo: Vec<i64> = vec![0; n];
  let mut leftmost: i64 = xs[0];
  for i in 1..n {
    if i == 1 {
      memo[i] = max(xs[0], xs[1]);
    } else if i % 2 == 0 {
      leftmost += xs[i];
      memo[i] = max(memo[i-2] + xs[i], memo[i-1]);
    } else {
      memo[i] = max(leftmost, memo[i-2] + xs[i]);
    }
  }
  println!("{}", memo[n-1]);
}