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)

#![allow(unused_imports, non_snake_case)]
use proconio::{
    input, fastout,
    marker::{Chars, Bytes, Usize1}
};

macro_rules! max {($ a : expr $ (, ) * ) => {{$ a } } ; ($ a : expr , $ b : expr $ (, ) * ) => {{std :: cmp :: max ($ a , $ b ) } } ; ($ a : expr , $ ($ rest : expr ) ,+ $ (, ) * ) => {{std :: cmp :: max ($ a , max ! ($ ($ rest ) ,+ ) ) } } ; }

#[fastout]
fn main() {
    input! {
        n: usize,
        a: [isize; n]
    }
    let mut dp = vec![vec![0; 3]; n + 4];
    for i in 0..n{
        dp[i + 4][0] = dp[i + 2][0] + a[i];
        dp[i + 4][1] = max!(dp[i+2][1], dp[i+1][0]) + a[i];
        if n & 1 > 0 {
            dp[i + 4][2] = max!(dp[i + 2][2], dp[i + 1][1], dp[i][0]) + a[i];
        }
    }
    if n & 1 > 0 {
        println!("{}", max!(dp[n+1][0], dp[n+2][1], dp[n+3][2]));
    } else {
        println!("{}", max!(dp[n+2][0], dp[n+3][1]));
    }
}