#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define Would
#define you
#define please

//世界は天才だなぁ

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);


	int N;
	cin >> N;
	int S[100000];
	rep(i, N) cin >> S[i];

	ll kotae = 0;
	rep1(ab, N / 2 - 1) {
		int n = ab;
		ll kari = 0;
		bool yoyuu = (N - 1) % n;
		while (n < N - 1 - ab && (yoyuu || n * 2 < N - 1)) {
			kari += S[n] + S[N - 1 - n];
			kotae = max(kotae, kari);
			n += ab;
		}
	}
	co(kotae);



	Would you please 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define Would
#define you
#define please

//世界は天才だなぁ

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);


	int N;
	cin >> N;
	int S[100000];
	rep(i, N) cin >> S[i];

	ll kotae = 0;
	rep1(ab, N / 2 - 1) {
		int n = ab;
		ll kari = 0;
		bool yoyuu = (N - 1) % n;
		while (n < N - 1 - ab && (yoyuu || n * 2 < N - 1)) {
			kari += S[n] + S[N - 1 - n];
			kotae = max(kotae, kari);
			n += ab;
		}
	}
	co(kotae);



	Would you please return 0;
}
```