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(
    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 = -1_000_000_000_000_000_000;
    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;
    if N & 1 == 0 {
        //偶数個の場合
        for i in (0..=N).step_by(2) {
            t = seven[i] + sodd[N] - sodd[i];
            ans = std::cmp::max(ans, t);
        }
    } else {
        //奇数個の場合
        let inf = -1_000_000_000_000_000_000;
        let mut max_over = vec![inf; N + 1];
        // 素直にやるとTLEするので、先に一部を求めておく。
        // max_over ...j >= i+1の時の、seven[N] + sodd[j] -seven[j] の最大値
        max_over[N] = sodd[N] - seven[N];
        let mut i = N - 2;
        loop {
            max_over[i] = std::cmp::max(max_over[i + 2], sodd[i] - seven[i]);
            if i >= 2 {
                i -= 2;
            } else {
                break;
            }
        }
        for i in (0..=N).step_by(2) {
            // for j in (i+1..=N).step_by(2) {
            //   t = seven[i]+ sodd[j] - sodd[i] + seven[N] - seven[j];
            t = seven[i] - sodd[i] + seven[N] + max_over[i + 1];
            ans = std::cmp::max(ans, t);
        }
    }
    println!("{}", ans);
}