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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class Main {
    InputStream is;

    int __t__ = 1;
    int __f__ = 0;
    int __FILE_DEBUG_FLAG__ = __f__;
    String __DEBUG_FILE_NAME__ = "src/C1";

    FastScanner in;
    PrintWriter out;

    public void solve() {
        int N = in.nextInt();
        long[] s = in.nextLongArray(N);

        long res = 0;
        int[] used = new int[N];
        for (int C = 1; C <= N - 2; C++) {
            int k = 1;
            long cur = 0;
            used[0] = C;
            while (true) {
                int ni1 = k * C, ni2 = N - 1 - k * C;
                if (!(0 <= ni1 && ni1 < N && 0 <= ni2 && ni2 < N)) {
                    break;
                }
                if (ni1 == ni2 || used[ni1] == C || used[ni2] == C) {
                    break;
                }
                used[ni1] = used[ni2] = C;
                cur += s[ni1];
                cur += s[ni2];
                res = Math.max(res, cur);
                k++;
            }
        }
        System.out.println(res);
    }

    /* MOD_CALCULATION */
    int MOD = 1_000_000_007;
    long ADD(long a, long b) {
        return (a + b) % MOD;
    }

    long SUB(long a, long b) {
        return (a - b + MOD) % MOD;
    }

    long MULT(long a, long b) {
        return (a * b) % MOD;
    }

    long POW(long a, long x) {
        long res = 1;
        for ( ; x > 0; x >>= 1) {
            if (x % 2 == 1) res = MULT(res, a);
            a = MULT(a, a);
        }

        return res;
    }

    long DIV(long a, long b) {
        return MULT(a, POW(b, MOD - 2));
    }
    /* end */


    public void run() {
        if (__FILE_DEBUG_FLAG__ == __t__) {
            try {
                is = new FileInputStream(__DEBUG_FILE_NAME__);
            } catch (FileNotFoundException e) {
                // TODO 自動生成された catch ブロック
                e.printStackTrace();
            }
            System.out.println("FILE_INPUT!");
        } else {
            is = System.in;
        }
        in = new FastScanner(is);
        out = new PrintWriter(System.out);

        solve();
    }

    public static void main(final String[] args) {
        new Main().run();
    }

    class FastScanner {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;

        public FastScanner(InputStream stream) {
            this.stream = stream;
            // stream = new FileInputStream(new File("dec.in"));

        }

        int read() {
            if (numChars == -1)
                throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        boolean isEndline(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] nextIntArray(int n) {
            int[] array = new int[n];
            for (int i = 0; i < n; i++)
                array[i] = nextInt();

            return array;
        }

        int[][] nextIntMap(int n, int m) {
            int[][] map = new int[n][m];
            for (int i = 0; i < n; i++) {
                map[i] = in.nextIntArray(m);
            }
            return map;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        long[] nextLongArray(int n) {
            long[] array = new long[n];
            for (int i = 0; i < n; i++)
                array[i] = nextLong();

            return array;
        }

        long[][] nextLongMap(int n, int m) {
            long[][] map = new long[n][m];
            for (int i = 0; i < n; i++) {
                map[i] = in.nextLongArray(m);
            }
            return map;
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        double[] nextDoubleArray(int n) {
            double[] array = new double[n];
            for (int i = 0; i < n; i++)
                array[i] = nextDouble();

            return array;
        }

        double[][] nextDoubleMap(int n, int m) {
            double[][] map = new double[n][m];
            for (int i = 0; i < n; i++) {
                map[i] = in.nextDoubleArray(m);
            }
            return map;
        }

        String next() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        String[] nextStringArray(int n) {
            String[] array = new String[n];
            for (int i = 0; i < n; i++)
                array[i] = next();

            return array;
        }

        String nextLine() {
            int c = read();
            while (isEndline(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndline(c));
            return res.toString();
        }
    }
}