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# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections;
using System.Collections.Generic;

namespace Main
{
    class Program
    {
        static void Main(string[] args)
        {
            //ここから
            //string[] str = Console.ReadLine().Split(' ');
            //int n = int.Parse(Console.ReadLine());
            
            int n = int.Parse(Console.ReadLine());
            string[] str = Console.ReadLine().Split(' ');

            var a = new long[n/2 + 1];
            var b = new long[n/2];

            for(int i = 0; i < n; i++)
            {
                if (i % 2 == 0) a[i/2] = int.Parse(str[i]);
                else b[i/2] = int.Parse(str[i]);
            }

            if (n % 2 == 0) { Console.WriteLine(Max(a.Sum(), b.Sum())); }
            else
            {
                long ans = Max(b.Sum(), a.Sum()-a.Min());
                //Console.WriteLine("奇数と偶数で見ると、{0}", ans);

                for(int i = 0; i < n / 2 + 1 ; i++)
                {
                    for (int p = i; p < n / 2; p++)
                    {
                        long temp = 0;
                        for (int j = 0; j < n / 2; j++)
                        {
                            if (j < i) temp += a[j];
                            else if (j < p) temp += b[j];
                            else temp += a[j + 1];
                        }
                        ans = Max(ans, temp);
                    }
                }

                for (int i = 0; i < n / 2 + 1 ; i++)
                {
                    long temp = 0;
                    for (int j = 0; j < n / 2; j++)
                    {
                        if (j < i) temp += b[j];
                        else temp += a[j + 1];
                    }
                    ans = Max(ans, temp);
                }

                Console.WriteLine(ans);
            }




            //ここまで
        }

        static long Max(long a, long b)
        {
            if ( a >= b) return a;
            else return b;
        }

        static long Min (long a, long b)
        {
            if (a <= b) return a;
            else return b;
        }

        

    }

}