#include <iostream>
#include <algorithm>
#include <vector>
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if(a < b){ a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if(a > b){ a = b; return 1; } return 0; }
typedef long long ll;
int main()
{
  int n; cin >> n;
  vector<ll> s(n);
  rep(i,n) cin >> s[i];

  vector<ll> a;
  rep(i,n) {
    int j = n - 1 - i;
    a.push_back(s[i] + s[j]);
  }    
  ll res = 0;
  rep1(i,n) {
    int idx = 0;
    ll sum = 0;
    int cr = n;
    if((n - 1) % i == 0) cr = n/2;
    while(idx < cr && i < n - idx - 1) {
      sum += a[idx];
      chmax(res, sum);
      idx += i;
    }
  }
  cout << res << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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 <algorithm>
#include <vector>
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if(a < b){ a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if(a > b){ a = b; return 1; } return 0; }
typedef long long ll;
int main()
{
  int n; cin >> n;
  vector<ll> s(n);
  rep(i,n) cin >> s[i];

  vector<ll> a;
  rep(i,n) {
    int j = n - 1 - i;
    a.push_back(s[i] + s[j]);
  }    
  ll res = 0;
  rep1(i,n) {
    int idx = 0;
    ll sum = 0;
    int cr = n;
    if((n - 1) % i == 0) cr = n/2;
    while(idx < cr && i < n - idx - 1) {
      sum += a[idx];
      chmax(res, sum);
      idx += i;
    }
  }
  cout << res << "\n";
  return 0;
}

```