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)

#![allow(unused_imports)]
#![allow(unused_macros)]
#![allow(non_snake_case)]
#![allow(dead_code)]
#![allow(unused_variables)]
#![allow(unused_mut)]
#![allow(unused_assignments)]
use proconio::input;
use proconio::marker::Usize1;
use std::collections::*;
use std::cmp::*;
use std::f64::consts::*;

const MOD: u64 = 1000000007;
const INF: i64 = std::i64::MAX / 4;

fn main() {
    input! {
        N: usize,
        A: [i64; N],
    }

    let (mut dp0, mut dp1, mut dp2) = (0, -INF, -INF);

    for (i, a) in A.iter().enumerate() {
        if i % 2 == 0 {
            dp1 = max(dp1, dp0);
            dp0 += a;
            dp2 += a;
        } else {
            dp2 = max(dp2, dp1);
            dp1 += a;
        }
    }

    if N % 2 == 0 {
        println!("{}", max(dp0, dp1));
    } else {
        println!("{}", max(dp1, dp2));
    }
}