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: Runtime Error)

#![allow(unused_imports)]
use std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use proconio::{input, marker::*};
use superslice::*;
fn main() {
    input! {
        n: usize,
        mut ai: [i64; n],
    }

    let mut ai2 = BinaryHeap::new();
    for (i, a) in ai.into_iter().enumerate() {
        ai2.push((a, i));
    }
    let mut ai = ai2;

    let m = n/2;
    let mut count = 0;
    let mut sum = 0;
    let mut sel = vec![false; n+2];

    while count < m {
        let (a, idx) = ai.pop().unwrap();
        let idx = idx + 1;
        if sel[idx-1] || sel[idx+1] {
            continue;
        }
        sel[idx] = true;
        count += 1;
        sum += a;
    }
    println!("{}", sum);
}