#include <cstdio>
#include <cstdint>
#include <vector>
#include <algorithm>
#include <utility>

int main() {
  size_t n;
  scanf("%zu", &n);

  std::vector<intmax_t> s(n);
  for (auto& si: s) scanf("%jd", &si);

  intmax_t res = 0;
  for (size_t c = 1; c <= n-1; ++c) {
    intmax_t cur = 0;
    std::vector<bool> reached(n);
    for (size_t kc = 0; kc <= n-1; kc += c) {
      if (reached[n-1-kc] || reached[kc]) break;
      if (kc == n-1-kc) break;
      reached[n-1-kc] = reached[kc] = true;
      if (n-1-kc < c) break;
      // fprintf(stderr, "+ [%zu] + [%zu]\n", kc, n-1-kc);
      cur += s[n-1-kc] + s[kc];
      res = std::max(res, cur);
    }
    // fprintf(stderr, "===\n");
  }
  printf("%jd\n", res);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 <cstdio>
#include <cstdint>
#include <vector>
#include <algorithm>
#include <utility>

int main() {
  size_t n;
  scanf("%zu", &n);

  std::vector<intmax_t> s(n);
  for (auto& si: s) scanf("%jd", &si);

  intmax_t res = 0;
  for (size_t c = 1; c <= n-1; ++c) {
    intmax_t cur = 0;
    std::vector<bool> reached(n);
    for (size_t kc = 0; kc <= n-1; kc += c) {
      if (reached[n-1-kc] || reached[kc]) break;
      if (kc == n-1-kc) break;
      reached[n-1-kc] = reached[kc] = true;
      if (n-1-kc < c) break;
      // fprintf(stderr, "+ [%zu] + [%zu]\n", kc, n-1-kc);
      cur += s[n-1-kc] + s[kc];
      res = std::max(res, cur);
    }
    // fprintf(stderr, "===\n");
  }
  printf("%jd\n", res);
}

```