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: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Numerics;
using System.Text;
using System.Text.RegularExpressions;
using System.Threading.Tasks;
using static System.Math;
using MethodImplAttribute = System.Runtime.CompilerServices.MethodImplAttribute;
using MethodImplOptions = System.Runtime.CompilerServices.MethodImplOptions;
public static class P
{
    public static void Main()
    {
        int n = int.Parse(Console.ReadLine());
        var a = Console.ReadLine().Split().Select(int.Parse).ToArray();
        var accumOdd = new long[n + 1];
        var accumEven = new long[n + 1];
        for (int i = 0; i < n; i++)
        {
            if (i % 2 == 0)
            {
                accumEven[i + 1] = accumEven[i] + a[i];
                accumOdd[i + 1] = accumOdd[i];
            }
            else
            {
                accumEven[i + 1] = accumEven[i];
                accumOdd[i + 1] = accumOdd[i] + a[i];
            }
        }
        if (n % 2 == 0)
        {
            var res = Max(accumOdd.Last(), accumEven.Last());
            var sum = 0L;
            for (int i = accumEven.Length - 3; i >= 0; i -= 2)
            {
                sum += a[i + 1];
                res = Max(res, accumEven[i] + sum);
            }
            Console.WriteLine(res);
            return;
        }
        else
        {
            //max(xoxoxox, oxoxoxx, xxoxoxo)
            var sum = accumEven.SkipLast(1).Last();
            var res = sum;
            var tailSum = 0L;
            var tailMax = 0L;
            for (int i = accumEven.Length - 4; i >= 0; i -= 2)
            {
                sum -= a[i]; sum += a[i + 1];
                tailSum += a[i + 2]; tailSum -= a[i + 1];
                tailMax = Max(tailMax, tailSum);
                res = Max(res, sum + tailMax);
            }
            Console.WriteLine(res);
            return;
        }
    }
}