#include<cstdio>

const int MAX_N = 100000;
int s[MAX_N];

int main() {
	int n;
	scanf("%d", &n);
	for(int i=0;i<n;i++) {
		scanf("%d", &s[i]);
	}
	long long max_score = 0;
	for(int ab=1;ab<=(n-2)/2;ab++) {
		int max_i = (n-2) / ab - 1;
		if((n-1) % ab == 0) {
			int tmp = ((n-1) / ab - 1) / 2;
			if(tmp < max_i) max_i = tmp;
		}
		long long score = 0;
		for(int i=1;i<=max_i;i++) {
			score += s[ab*i] + s[n-1-ab*i];
			if(score > max_score) max_score = score;
		}
	}
	printf("%lld\n", max_score);
	return 0;
}

### 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>

const int MAX_N = 100000;
int s[MAX_N];

int main() {
	int n;
	scanf("%d", &n);
	for(int i=0;i<n;i++) {
		scanf("%d", &s[i]);
	}
	long long max_score = 0;
	for(int ab=1;ab<=(n-2)/2;ab++) {
		int max_i = (n-2) / ab - 1;
		if((n-1) % ab == 0) {
			int tmp = ((n-1) / ab - 1) / 2;
			if(tmp < max_i) max_i = tmp;
		}
		long long score = 0;
		for(int i=1;i<=max_i;i++) {
			score += s[ab*i] + s[n-1-ab*i];
			if(score > max_score) max_score = score;
		}
	}
	printf("%lld\n", max_score);
	return 0;
}

```