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)

#![allow(unused_imports)]
use petgraph::unionfind::UnionFind;
use proconio::*;
use std::cmp::*;
use std::fs;
use std::io::*;

const INF: i64 = 100100100100;
const MAX_N: usize = 200005;

#[fastout]
fn main() {
   input! {
      n: usize,
      a: [i64; n]
   }
   let k = (n % 2) + 1;
   let mut dp = [[-INF; 4]; MAX_N+1];
   dp[0][0] = 0;
   for i in 0..n {
      for j in 0..(k+1) {
         dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j]);
         dp[i+1][j] = max(dp[i+1][j], match (i+j)%2 {
             0 => dp[i][j] + a[i],
             _ => dp[i][j],
         });
      }
   }
   println!("{}", dp[n][k]);
}