#include <iostream>
#include <vector>
#include <cmath>

using lli=long long int;

int main(){
  int N;
  std::cin >> N;

  std::vector<lli> s_arr(N);
  for(int i = 0; i < N; i++){
    std::cin >> s_arr[i];
  }

  lli score_max = -1;
  for(int nforward = 1; nforward < N; nforward++){
    lli score = s_arr[N-1];
    lli score_n = score;
    int nstep = 0;
    while(true){
      nstep++;
      int a = N-1-nforward*nstep;
      if(a%nforward==0 && 2*a<=N-1) break;
      if(a<nforward) break;
      score += s_arr[nstep*nforward]+s_arr[a];
      score_n = std::max(score_n, score);
    }
    score_max = std::max(score_max, score_n);
  }

  std::cout << score_max << std::endl;

  return 0;
}


### Prompt

Generate a cpp solution to the following problem:

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.

Examples

Input

5
0 2 5 1 0


Output

3


Input

6
0 10 -7 -4 -13 0


Output

0


Input

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


Output

59

### Response

```cpp
#include <iostream>
#include <vector>
#include <cmath>

using lli=long long int;

int main(){
  int N;
  std::cin >> N;

  std::vector<lli> s_arr(N);
  for(int i = 0; i < N; i++){
    std::cin >> s_arr[i];
  }

  lli score_max = -1;
  for(int nforward = 1; nforward < N; nforward++){
    lli score = s_arr[N-1];
    lli score_n = score;
    int nstep = 0;
    while(true){
      nstep++;
      int a = N-1-nforward*nstep;
      if(a%nforward==0 && 2*a<=N-1) break;
      if(a<nforward) break;
      score += s_arr[nstep*nforward]+s_arr[a];
      score_n = std::max(score_n, score);
    }
    score_max = std::max(score_max, score_n);
  }

  std::cout << score_max << std::endl;

  return 0;
}


```