#include <iostream>
#include <string>
#include <utility>
#include <vector>
#include <algorithm>
#include <set>
#define llint long long

using namespace std;
typedef pair<llint, llint> P;

llint n;
llint a[100005];

int main(void)
{
	cin >> n;
	for(int i = 0; i < n; i++) cin >> a[i];
	
	llint ans = 0;
	for(int i = 1; i < n-1; i++){
		llint sum = 0;
		if(i >= n-1-i) continue;
		for(int j = i; j <= n-1; j+=i){
			if((n-1-j)%i == 0 && j >= (n-1-j)) break;
			if((n-1-j)<=i) break;
			sum += a[j] + a[n-1-j];
			ans = max(ans, sum);
			//cout << i << " " << j << " " << ans << endl;
		}
	}
	cout << ans << endl;
	
	return 0;
}
### Prompt

In CPP, your task is to solve 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 <string>
#include <utility>
#include <vector>
#include <algorithm>
#include <set>
#define llint long long

using namespace std;
typedef pair<llint, llint> P;

llint n;
llint a[100005];

int main(void)
{
	cin >> n;
	for(int i = 0; i < n; i++) cin >> a[i];
	
	llint ans = 0;
	for(int i = 1; i < n-1; i++){
		llint sum = 0;
		if(i >= n-1-i) continue;
		for(int j = i; j <= n-1; j+=i){
			if((n-1-j)%i == 0 && j >= (n-1-j)) break;
			if((n-1-j)<=i) break;
			sum += a[j] + a[n-1-j];
			ans = max(ans, sum);
			//cout << i << " " << j << " " << ans << endl;
		}
	}
	cout << ans << endl;
	
	return 0;
}
```