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)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

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

  let inf = 1_i64 << 60;
  let m = n / 2;
  let mut dp = vec![vec![vec![-inf; 3]; m + 1]; n + 1];
  dp[0][0][0] = 0;
  dp[0][0][1] = 0;
  dp[0][0][2] = 0;
  for i in 0..n {
    let r = (i + 1) / 2;
    let l = if r > 0 { r - 1 } else { 0 };
    for j in l..=r {
      // select
      if j + 1 <= m {
        dp[i + 1][j + 1][0] = vec![
          dp[i + 1][j + 1][0],
          dp[i][j][1] + av[i],
          dp[i][j][2] + av[i],
        ]
        .into_iter()
        .max()
        .unwrap();
      }
      // not select
      dp[i + 1][j][1] = vec![dp[i + 1][j][1], dp[i][j][0]]
        .into_iter()
        .max()
        .unwrap();
      dp[i + 1][j][2] = vec![dp[i + 1][j][2], dp[i][j][1]]
        .into_iter()
        .max()
        .unwrap();
    }
  }
  println!("{}", dp[n][m].clone().into_iter().max().unwrap());
}