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::input;

fn main() {
    input! {
     n:usize,
     mut a:[isize;n]
    }
    a.insert(0, 0);
    a.insert(0, 0);
    let n = n + 2;

    let mut dp = vec![vec![-999999999999_isize; 3]; n + 2];
    dp[0][0] = 0;
    dp[1][0] = 0;
    for i in 0..n {
        dp[i + 2][0] = dp[i][0] + a[i];
        if i <= 1 {
            continue;
        }
        dp[i + 2][1] = std::cmp::max(dp[i + 1][0], dp[i][1] + a[i]);
        dp[i + 2][2] = std::cmp::max(dp[i + 1][1], dp[i][2] + a[i]);
    }
    let mut ans = std::cmp::max(dp[n + 1][0], dp[n + 1][1]);
    if n % 2 != 0 {
        ans = std::cmp::max(ans, dp[n + 1][2]);
    }
    //dbg!(dp);
    println!("{}", ans);
}