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;

fn main() {
    input! {
        n :usize,
        av: [i64; n],
    };
    let inf = 1_000_000_000_000_000_i64;
    let m = if n % 2 == 0 { 1 } else { 2 };
    // dp[i][j] = x
    // i (0<=i<=n) までで j (0<=j<=m) 個を飛ばしたときの最大値が x
    let mut dp = vec![vec![-inf; m + 1]; n + 1];
    dp[0][0] = 0;
    for (i, &a) in av.iter().enumerate() {
        for j in 0..=m {
            // 飛ばす場合
            if j + 1 <= m {
                dp[i + 1][j + 1] = std::cmp::max(dp[i + 1][j + 1], dp[i][j]);
            }
            // 飛ばさない場合
            dp[i + 1][j] = std::cmp::max(
                dp[i + 1][j],
                dp[i][j] + if (i + j) % 2 == 0 { a } else { 0 },
            );
        }
    }
    let ans = dp[n][m];
    println!("{}", ans);
}