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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rep1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define irep(i, a, n) for (int i = a; i < (int)(n); ++i)
#define rrep(i, n) for (int i = (int)(n)-1; i >= 0; --i)
#define rrep1(i, n) for (int i = (int)(n); i >= 1; --i)
#define allrep(V, v) for (auto &&V : v)
#define all(x) (x).begin(), (x).end()
const int INF = 1 << 30;
const long long INFL = 1LL << 62;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
using lint = long long;

using namespace std;

int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;
  vector<int> s(n);
  lint maxsum = -INFL;
  rep(i, n) cin >> s[i];
  int n1 = n - 1;
  rep1(c, n1 - 1) {
    set<int> alr;
    lint sum = 0;
    for (int nc = 0; nc < n1; nc += c) {
      int a = n1 - nc;
      int b = a - c;
      if (a <= b || b < 0) continue;
      if (alr.find(nc) != alr.end()) break;
      alr.insert(nc);
      if (alr.find(n1 - nc) != alr.end()) break;
      alr.insert(n1 - nc);
      sum += s[nc] + s[n1 - nc];
      maxsum = max(maxsum, sum);
    }
  }
  cout << maxsum << endl;
  return 0;
}