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 num::integer::gcd;
use proconio::*;
use std::cmp::*;
use std::collections::VecDeque;
use std::ops::*;

const INF: i64 = 1001001001001001001;

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

   let k = n % 2 + 1;
   let mut dp = vec![vec![-INF; 4]; n + 1];
   dp[0][0] = 0;
   for i in 0..n {
      for j in 0..=k {
         dp[i + 1][j + 1] = dp[i][j];
         dp[i + 1][j] = max(
            dp[i + 1][j],
            dp[i][j] + if (i + j) % 2 == 0 { a[i] } else { 0 },
         );
      }
   }
   println!("{}", dp[n][k]);
}