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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.InputStream;
import java.io.PrintStream;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws Exception {
        solve(System.in, System.out);
    }

    static void solve(InputStream is, PrintStream os) {
        // Your code here!
        Scanner scan = new Scanner(is);
        int N = scan.nextInt();
        int A[] = new int[N];

        for (int i = 0; i < N; i++) {
            A[i] = scan.nextInt();
        }

        if (N == 2) {
            os.println(Math.max(A[0], A[1]));
            return;
        }

        long zero[] = new long[N];
        long one[] = new long[N];
        long two[] = new long[N];

        int upperZero = 0;
        int upperOne = 0;
        if (N % 2 == 0){
            upperZero = N-1;
            upperOne = N;
        } else {
            upperZero = N-2;
            upperOne = N-1;
        }


        zero[0] = A[0];
        for(int i = 2; i < upperZero; i += 2) {
            zero[i] += zero[i-2] + A[i];
        }
        one[1] = A[1];
        for(int i = 3; i < upperOne; i += 2) {
            one[i] += Math.max(zero[i-3] + A[i], one[i-2] + A[i]);
        }

        two[2] = A[2];
        for(int i = 4; i < N; i += 2) {
            two[i] += Math.max(Math.max(A[i]+one[i-3], A[i]+zero[i-4]), two[i-2]+A[i]);
        }

        if(N%2 != 0)
            os.println(Math.max(Math.max(two[N-1], one[N-2]), zero[N-3]));
        else
            os.println(Math.max(one[N-1], zero[N-2]));
    }
}