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 proconio::input;
use std::cmp;

fn main() {
    input! {
        n : usize,
        a : [i64; n]
    };
    let mut dp = vec![[-(1 << 50); 3]; n];
    for i in 0..cmp::min(n, 3) {
        dp[i][i] = a[i];
    }
    for i in 0..n {
        for j in 0..3 {
            for k in 0..3 {
                if i+k+2 >= n || j+k >= 3 { break;}
                dp[i+k+2][j+k] = cmp::max(dp[i+k+2][j+k], dp[i][j] + a[i+k+2]);
            }
        }
    }
    let res = if n%2 == 1 {
        cmp::max(dp[n-1][2], cmp::max(dp[n-2][1], dp[n-3][0]))
    } else {
        cmp::max(dp[n-1][1], dp[n-2][0])
    };
    println!("{}", res);
}