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 C# (Mono-mcs 6.8.0.105) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
public class Test
{

    public static void Main()
    {

        var n = int.Parse(Console.ReadLine());
        int[] arr = new int[n];

        string[] input = Console.ReadLine().Split(' ');

        for (int i = 0; i < n; i++)
        {
            arr[i] = int.Parse(input[i]);
        }


        Console.Write(MaxAlSum(arr, n));
    }

    static long MaxAlSum(int[] arr, int n)
    {
        long max1 = 0;
        long max2 = 0;
        long[] dp = new long[n];
        dp[0] = arr[0];
        for (int i = 2; i < n; i = i + 2)
        {
            dp[i] = Math.Max(dp[i - 2] + arr[i], arr[i]);
        }
        max1 = dp[n - 2];
        long[] dp1 = new long[n];
        dp1[1] = arr[1];
        for (int i = 3; i < n; i = i + 2)
        {
            dp1[i] = Math.Max(dp1[i - 2] + arr[i], arr[i]);
        }
        max2 = dp1[n - 1];

        return Math.Max(max1, max2);
    }
}