Problem:
Score : 600 points

Problem Statement

There is an infinitely large pond, which we consider as a number line.
In this pond, there are N lotuses floating at coordinates 0, 1, 2, ..., N-2 and N-1.
On the lotus at coordinate i, an integer s_i is written.

You are standing on the lotus at coordinate 0. You will play a game that proceeds as follows:

1. Choose positive integers A and B. Your score is initially 0.

2. Let x be your current coordinate, and y = x+A. The lotus at coordinate x disappears, and you move to coordinate y.

If y = N-1, the game ends.

If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.

If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.

3. Let x be your current coordinate, and y = x-B. The lotus at coordinate x disappears, and you move to coordinate y.

If y = N-1, the game ends.

If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.

If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.

4. Go back to step 2.

You want to end the game with as high a score as possible.
What is the score obtained by the optimal choice of A and B?

Constraints

3 \leq N \leq 10^5

-10^9 \leq s_i \leq 10^9

s_0=s_{N-1}=0

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
s_0 s_1 ...... s_{N-1}

Output

Print the score obtained by the optimal choice of A and B.

Sample Input 1

5
0 2 5 1 0

Sample Output 1

3

If you choose A = 3 and B = 2, the game proceeds as follows:

Move to coordinate 0 + 3 = 3. Your score increases by s_3 = 1.

Move to coordinate 3 - 2 = 1. Your score increases by s_1 = 2.

Move to coordinate 1 + 3 = 4. The game ends with a score of 3.

There is no way to end the game with a score of 4 or higher, so the answer is 3. Note that you cannot land the lotus at coordinate 2 without drowning later.

Sample Input 2

6
0 10 -7 -4 -13 0

Sample Output 2

0

The optimal strategy here is to land the final lotus immediately by choosing A = 5 (the value of B does not matter).

Sample Input 3

11
0 -4 0 -99 31 14 -15 -39 43 18 0

Sample Output 3

59

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Text;
using static System.Math;

class Program
{
    static int n;
    static long[] s;

    static void Main()
    {
        n = int.Parse(Console.ReadLine());
        s = Console.ReadLine().Split().Select(long.Parse).ToArray();
        long ans = 0;
        for (int c = 1; c < n - 1; c++) // 各c=a-bについて
        {
            long[] sum1 = new long[(n - 1) / c + 1]; // 0,c,2c...の累積和
            sum1[0] = s[0];
            for (int i = 1; i < sum1.Length; i++)
                sum1[i] = sum1[i - 1] + s[i * c];
            long[] sum2 = new long[(n - 1) / c + 1]; // n-1,n-1-c,n-1-2c,...の累積和
            sum2[0] = s[n - 1];
            for (int i = 1; i < sum2.Length; i++)
                sum2[i] = sum2[i - 1] + s[n - 1 - i * c];
            long[] scores = new long[(n - 1) / c + 1];
            for (int i = 0; i < scores.Length; i++)
            {
                if ((n - 1) % c == 0 && (n - 1) / c <= 2 * i) continue;
                scores[i] = sum1[i] + sum2[i];
            }
            ans = Max(ans, scores.Max());
        }
        Console.WriteLine(ans);
    }
}