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(
    non_snake_case,
    unused_variables,
    unused_assignments,
    unused_mut,
    unused_imports,
    dead_code
)]
//use permutohedron::LexicalPermutation as _;
use proconio::input;
//use proconio::marker::*;
//use std::collections::*;
use std::cmp::*;
// use std::num;
fn main() {
    input! {N:usize, A:[i64;N]}
    let inf = std::i64::MIN / 2;
    let mut ans = inf;
    let mut dp = vec![vec![vec![inf; 2]; 2]; N];
    dp[0][0][1] = A[0];
    dp[1][0][0] = A[0];
    dp[1][0][1] = A[1];
    // dp[i][j] i番目までのうち、((i+1)/2)-j 個選んでいるときの最大値。
    for i in 2..N {
        let k = (i + 1) / 2;
        if i & 1 == 1 {
            // 奇数番目(偶数個)で、半分選び、i番目を選ばないケース
            dp[i][0][0] = dp[i - 1][0][1];
            // 奇数番目(偶数個)で、半分選び、i番目を選ぶケース
            dp[i][0][1] = dp[i - 1][1][0] + A[i];
            // 奇数番目(偶数個)で、半分-1個選び、i番目を選ばないケース
            dp[i][1][0] = max(dp[i - 1][1][0], dp[i - 1][1][1]);
        // 奇数番目(偶数個)で、半分-1個選び、i番目を選ぶケース
        // その後条件を満たさないので考慮扶養。
        } else {
            // 偶数番目目(奇数個)
            // (i+1)/2 個選び、i番目を選ばないケース
            // 無理
            // (i+1)/2 個選び、i番目を選ぶケース
            dp[i][0][1] = dp[i - 1][0][0] + A[i];
            // (i+1)/2-1 個選び、i番目を選ばないケース
            dp[i][1][0] = max(dp[i - 1][0][0], dp[i - 1][0][1]);
            // (i+1)/2-1 個選び、i番目を選ぶケース
            dp[i][1][1] = dp[i - 1][1][0] + A[i];
            if i == 2 {
                dp[i][1][1] = A[i];
            }
            // (i+1)/2-2 個選び、i番目を選ばないケース
            // dp[i][2][0] = max(dp[i - 1][1][0], dp[i - 1][1][1]);
            // (i+1)/2-2 個選び、i番目を選ぶケース
            // 無理 dp[i][2][1] = dp[i - 1][2][0] + A[i];
        }
    }
    //println!("{:?}", dp);
    if N & 1 == 0 {
        ans = max(dp[N - 1][0][0], dp[N - 1][0][1]);
    } else {
        ans = max(dp[N - 1][1][0], dp[N - 1][1][1]);
    }

    println!("{}", ans);
}