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(
    non_snake_case,
    unused_variables,
    unused_assignments,
    unused_mut,
    unused_imports,
    dead_code
)]
//use permutohedron::LexicalPermutation as _;
use proconio::input;
//use proconio::marker::*;
//use std::collections::*;
//use std::cmp::*;
// use std::num;
fn main() {
    input! {N:usize, A:[i64;N]}
    let mut ans = 0;
    let mut sodd = vec![0i64; N + 1];
    let mut seven = vec![0i64; N + 1];
    sodd[0] = 0;
    for i in 0..N {
        if i & 1 == 1 {
            sodd[i + 1] = sodd[i] + A[i];
            seven[i + 1] = seven[i];
        } else {
            sodd[i + 1] = sodd[i];
            seven[i + 1] = seven[i] + A[i];
        }
    }
    // 連続する箇所をi,jと置く。
    let mut t = 0;
    //println!("sodd {:?}", sodd);
    //println!("seve {:?}", seven);

    if N & 1 == 0 {
        //偶数個の場合
        for i in 1..=N / 2 {
            t = seven[2 * i - 2] + sodd[N] - sodd[2 * i - 1];
            ans = std::cmp::max(ans, t);
        }
    } else {
        //奇数個の場合
        for i in 1..=N / 2 {
            for j in i..=N / 2 {
                //println!("i{},j{},seven={}", i, j, seven[2 * i - 2]);
                //println!("sodd={}-{}", sodd[2 * j], sodd[2 * i - 1]);
                //println!("seven={}-{}", seven[N], seven[2 * j + 1]);
                t = seven[2 * i - 2];
                t += sodd[2 * j] - sodd[2 * i - 1];
                t += seven[N] - seven[2 * j + 1];
                ans = std::cmp::max(ans, t);
            }
        }
    }
    println!("{}", ans);
}