Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

5
0 2 5 1 0

Sample Output 1

3

If you choose A = 3 and B = 2, the game proceeds as follows:

Move to coordinate 0 + 3 = 3. Your score increases by s_3 = 1.

Move to coordinate 3 - 2 = 1. Your score increases by s_1 = 2.

Move to coordinate 1 + 3 = 4. The game ends with a score of 3.

There is no way to end the game with a score of 4 or higher, so the answer is 3. Note that you cannot land the lotus at coordinate 2 without drowning later.

Sample Input 2

6
0 10 -7 -4 -13 0

Sample Output 2

0

The optimal strategy here is to land the final lotus immediately by choosing A = 5 (the value of B does not matter).

Sample Input 3

11
0 -4 0 -99 31 14 -15 -39 43 18 0

Sample Output 3

59

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define int long long
#define pb push_back

using namespace std;

const int N=1e5+5, INF=1e18;

int n, arr[N], mx=-INF;
signed main()
{
	cin >> n;
	for(int i = 0; i < n; i++)
	{
		cin >> arr[i];
	}
	for(int c = 1; c < n; c++)
	{
		int currmx=-INF;
		vector<int> jc, jcpd, pjc, sjcpd;
		int d = c + (n-1)%c, sz = 0;
		while(n+c-1>2*d) d+=c;
	//	cout << c << " - " << d << endl;
		jc.pb(0); jcpd.pb(0);	pjc.pb(0); sjcpd.pb(0);
		for(int j = d; j < n; j+=c)
		{
			sz++;
			if(j!=n-1) jc.pb(arr[sz*c]);
			else jc.pb(0);
			jcpd.pb(arr[j]);
			pjc.pb(pjc.back()+jc.back());
			sjcpd.pb(0);
		}
		sjcpd[sz]=jcpd[sz];
		for(int j = sz - 1; j >= 0; j--)
		{
			sjcpd[j]=sjcpd[j+1]+jcpd[j];
		}
	/*	for(int j = 0; j <= sz; j++)
		{
			cout << jc[j] << "  ";
		}
		cout << endl;
		for(int j = 0; j <= sz; j++)
		{
			cout << jcpd[j] << "  ";
		}
		cout << endl;*/
		for(int j = 0; j <= sz; j++)
		{
//			cout << j << " - " << sjcpd[j] << " " << pjc[sz - j] << endl;
			currmx = max(currmx, sjcpd[j] + pjc[sz - j]);
		}
		mx = max(currmx, mx);
	}
	cout << mx;
}