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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use competitive::prelude::*;


#[allow(unused)]
macro_rules! max {
    ($x:expr) => ( $x );
    ($x:expr, $($xs:expr),+) => {
        {
            std::cmp::max($x, max!( $($xs),+ ))
        }
    };
}

fn main() {
    input! {
        n: usize,
        a: [i64; n],
    }
    // dp[i][j] = i回ジャンプ, 右端がa[j]のときの最大値
    let ans =
        if n % 2 == 0 {
            // ジャンプは1回以下
            let mut dp = vec![vec![-1e10 as i64; n]; 2];
            dp[0][0] = a[0];
            dp[1][1] = a[1];
            for i in 3..n {
                dp[0][i] = dp[0][i - 2] + a[i];
                dp[1][i] = a[i] + max!(dp[0][i - 3], dp[1][i - 2]);
            }
            max!(
                dp[0][n - 2],
                dp[1][n - 1]
            )
        } else {
            // ジャンプは2回以下
            let mut dp = vec![vec![-1e10 as i64; n]; 3];
            dp[0][0] = a[0];
            dp[0][2] = a[0] + a[2];
            dp[1][1] = a[1];
            dp[1][3] = a[3] + max!(dp[0][0], dp[1][1]);
            dp[2][2] = a[2];
            for i in 4..n {
                dp[0][i] = a[i] + dp[0][i - 2];
                dp[1][i] = a[i] + max!(dp[0][i - 3], dp[1][i - 2]);
                dp[2][i] = a[i] + max!(dp[0][i - 4], dp[1][i - 3], dp[2][i - 2]);
            }
            max!(
                dp[0][n - 3],
                dp[1][n - 2],
                dp[2][n - 1]
            )
        };
    println!("{}", ans);
}
*/

fn main() {
    let exe = "/tmp/bin40803E68";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";