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)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String {
            bytes
                .by_ref()
                .map(|r| r.unwrap() as char)
                .skip_while(|c| c.is_whitespace())
                .take_while(|c| !c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
        }
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_imports)]
use std::cmp::{min, max};

fn main() {
    input!{
        n: usize,
        v: [i64; n]
    }
    if n % 2 == 0 {
        let mut e = 0;
        let mut o = 0;
        for i in 0..n {
            if i % 2 == 0 {
                e += v[i];
            } else {
                o += v[i];
            }
        }
        println!("{}", max(e, o));
        return;
    }
    if n == 3 {
        println!("{}", max(v[0], max(v[1], v[2])));
        return;
    }
    let mut dp = vec![[i64::min_value(); 3]; n];
    dp[0][0] = v[0];
    dp[1][1] = v[1];
    dp[2][0] = v[0] + v[2];
    dp[2][2] = v[2];
    dp[3][1] = max(v[1], v[0]) + v[3];
    for i in 4..n {
        for j in 0..3 {
            if dp[i - 2][j] != i64::min_value() {
                dp[i][j] = dp[i - 2][j] + v[i];
            }
        }
        if dp[i - 3][0] != i64::min_value() {
            dp[i][1] = max(dp[i - 3][0] + v[i], dp[i][1]);
        }
        if dp[i - 3][1] != i64::min_value() {
            dp[i][2] = max(dp[i - 3][1] + v[i], dp[i][2]);
        }
        if dp[i - 4][0] != i64::min_value() {
            dp[i][2] = max(dp[i - 4][0] + v[i], dp[i][2]);
        }
    }
    println!("{}", max(dp[n - 1][2], max(dp[n - 2][1], dp[n - 3][0])));
}