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::cmp::*;
use std::collections::*;
// use std::num;
fn main() {
    input! {N:usize, A:[i64;N]}
    let mut ans = 0;
    let mut dp = vec![HashMap::new(); N];
    // i 番目までの数のうち(i+1)/2 -j 個選んだ時の数をdp[i][j]で表す。
    dp[0].insert(0, 0);
    dp[0].insert(1, A[0]);
    dp[1].insert(0, 0);
    dp[1].insert(1, max(A[0], A[1]));
    for i in 2..N {
        for j in 0..3 {
            let k = (i + 1) / 2 - j;
            if k < 1 {
                break;
            }
            let x = *dp[i - 1].get(&k).unwrap_or(&0);
            let y = *dp[i - 2].get(&(k - 1)).unwrap_or(&0);
            dp[i].insert(k, max(x, y + A[i]));
        }
    }
    ans = *dp[N - 1].get(&(N / 2)).unwrap();
    println!("{}", ans);
}