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)

#[allow(unused_imports)]
use std::cmp;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use proconio::derive_readable;
use proconio::fastout;
#[allow(unused_imports)]
use num::Integer;
#[allow(unused_imports)]
use itertools::Itertools;

#[fastout]
fn main() {
  input! {
    n: usize,
    a: [i64; n],
  }
  let mut sum = vec![0; n+2];
  for (i, &x) in a.iter().enumerate() {
    sum[i+2] = sum[i] + x;
  }
  if n % 2 == 0 {
    let mut ans = - (1i64 << 60);
    let mut i = 0;
    while i <= n {
      ans = cmp::max(ans, sum[i] + sum[n+1] - sum[i+1]);
      i += 2;
    }
    println!("{}", ans);
  } else {
    let mut ans = - (1i64 << 60);
    let mut right = vec![- (1i64 << 60); n+3];
    let mut i = n;
    loop {
      right[i] = cmp::max(right[i+2], sum[i] + sum[n+1] - sum[i+1]);
      if i < 2 {
        break;
      }
      i -= 2;
    }
    let mut j = 0;
    while j < n {
      ans = cmp::max(ans, right[j+1] - sum[j+1] + sum[j]);
      j += 2;
    }
    println!("{}", ans);
  }
}