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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <bitset>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <cstring>
#include <utility>
#include <vector>
#include <complex>
#include <valarray>
#include <fstream>
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <numeric>
#include <climits>
#include <random>

#define _overload(a, b, c, d, ...) d
#define _rep1(X, A, Y) for (int (X) = (A);(X) <= (Y);++(X))
#define _rep2(X, Y) for (int (X) = 0;(X) < (Y);++(X))
#define rep(...) _overload(__VA_ARGS__, _rep1, _rep2)(__VA_ARGS__)
#define rrep(X,Y) for (int (X) = Y-1;(X) >= 0;--(X))
#define all(X) (X).begin(),(X).end()
#define len(X) ((int)(X).size())
#define mod(n, m) (((n)%(m)+(m))%m)
#define fi first
#define sc second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> Pii;
typedef pair<ll,ll> Pll;
const int INFINT = 1 << 30;                          // 1.07x10^ 9
const ll INFLL = 1LL << 60;                          // 1.15x10^18
const double EPS = 1e-10;
const int MOD = 1000000007;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

const int MAX_N = 100000;
int N;
ll S[MAX_N];

int main() {
  std::ios::sync_with_stdio(false);std::cin.tie(0);
  
  cin >> N;
  rep(i, N) cin >> S[i];

  ll ans = 0;
  for (int C = 1; C < N; ++C) {

    ll tmp = 0, ma = 0;
    int lb = 0, ub = N-1;
    while (lb <= N-1 && ub >= 0) {
      if (lb == ub || (lb-ub) == C) break;
      tmp += S[ub] + S[lb];
      ma = max(ma, tmp);
      lb += C; ub -= C;
    }

    ans = max(ans, ma);
  }

  cout << ans << endl;

  return 0;
}