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 inf = std::i64::MIN / 4;
    let mut ans = inf;
    let mut dp = vec![vec![inf; 3]; N + 1];
    // dp[i][j] i個まで進めた時、スキップ消費済みの数をjとする。
    // Nが奇数なら、最終的にスキップを1個または2個消費する
    // Nが偶数なら、最終的にスキップを0個または1個消費する
    // スキップを3個以上消費することはできない。
    let k = N % 2 + 1;
    dp[0][0] = 0;
    for i in 0..N {
        if i % 2 == 0 {
            // 偶数個 なら、奇数番目は…。
            dp[i + 1][0] = dp[i][0] + A[i];
            dp[i + 1][1] = max(dp[i][0], dp[i][1]);
            dp[i + 1][2] = dp[i][2] + A[i];
        } else {
            dp[i + 1][0] = dp[i][0];
            dp[i + 1][1] = dp[i][1] + A[i];
            dp[i + 1][2] = max(dp[i][1], dp[i][2]);
        }
    }
    //println!("{:?}", dp);
    let ans = max(dp[N][k - 1], dp[N][k]);
    println!("{}", ans);
}