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;
use std::cmp::max;

fn main() {
    input! {
        n: usize,
        a: [isize; n],
    }

    let k = n / 2;

    let mut dp = vec![vec![std::isize::MIN / 2; k + 1]; n];
    dp[0][0] = 0;
    dp[1][0] = 0;
    dp[0][1] = a[0];
    dp[1][1] = max(a[0], a[1]);

    for i in 2..n {
        for j in 1..=k {
            if i / 2 - 1 <= j && j <= (i + 1) / 2 {
                dp[i][j] = max(a[i] + dp[i - 2][j - 1], dp[i - 1][j])
            }
        }
    }

    println!("{}", dp[n - 1][k]);
}