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)

use proconio::input;

#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars, Usize1};
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::ops::*;

#[derive(Clone, Debug, Default)]
struct Struct;

//#[proconio::fastout]
fn main() {
    input! {
        n: usize,
        a: [i64;n],
    }

    /* _dp0[i][j] : [0,i) から j 個選んで、a[i-1]を選ばなかった場合
       _dp1[i][j] : [0,i) から j 個選んで、a[i-1]を選んだ場合
       (i-1)/2 <= j <= (i+1)/2

       (i-1)/2 = b とする
       dp0[i][j-b+1] = dp0[i][j]
       dp1[i][j-b+1] = dp1[i][j]
    */

    const MIN_SUM: i64 = -1000_000_000i64 * 200000;

    let mut dp0 = vec![vec![MIN_SUM; 4]; n + 1];
    let mut dp1 = vec![vec![MIN_SUM; 4]; n + 1];
    fn f(i:usize,j:i64)->usize{
        let b =  (i as i64 - 1) >> 1;
        ( j + 1 - b ) as usize
    }

    dp0[0][f(0,0)] = 0;
    for i in 1..=n {
        let b = (i as i64 - 1) >> 1;
        for j in b..=b + 1 {
            dp0[i][f(i,j)] = max(dp0[i - 1][f(i-1,j)], dp1[i - 1][f(i-1,j)]);
            dp1[i][f(i,j)] = dp0[i - 1][f(i-1,j) - 1] + a[i - 1];
            //println!("{} {}", f(i,j), f(i-1,j));
            //println!("{} {}", dp0[i][f(i,j)], dp1[i][f(i,j)]);
        }
    }
    let j = n as i64 / 2;
    println!("{}", max(dp0[n][f(n,j)], dp1[n][f(n,j)]));
}