#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
int main(){
	int N; long long int S[100010] = {}; 
	cin >> N; 
	for(int i = 0; i < N; i++){
		cin >> S[i]; 
	}
	long long int ans = 0; // m=A-Bとするとmは1以上N-1以下
	for(int m = 1; m < N; m++){
		long long int mx = 0; 
		long long int tmp = 0; // S[0] + S[N-1] から始めて次は S[m] + S[N-1-m] を足す感じ
		for(int j = 1; N - 1 - j * m > m; j++){ // B > 0 に注意！
			if((N - 1) % m == 0 && N - 1 - j * m <= j * m) break; 
			tmp += S[j * m] + S[N - 1 - j * m]; 
			mx = max(mx, tmp); 
		}
		ans = max(ans, mx); 
	}
	cout << ans << endl; 
}
### 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 <bits/stdc++.h>
#include <algorithm>
using namespace std;
int main(){
	int N; long long int S[100010] = {}; 
	cin >> N; 
	for(int i = 0; i < N; i++){
		cin >> S[i]; 
	}
	long long int ans = 0; // m=A-Bとするとmは1以上N-1以下
	for(int m = 1; m < N; m++){
		long long int mx = 0; 
		long long int tmp = 0; // S[0] + S[N-1] から始めて次は S[m] + S[N-1-m] を足す感じ
		for(int j = 1; N - 1 - j * m > m; j++){ // B > 0 に注意！
			if((N - 1) % m == 0 && N - 1 - j * m <= j * m) break; 
			tmp += S[j * m] + S[N - 1 - j * m]; 
			mx = max(mx, tmp); 
		}
		ans = max(ans, mx); 
	}
	cout << ans << endl; 
}
```