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)

// #includes {{{
#ifdef MY_DEBUG
#include "header/header.hpp"
#else
#include <bits/stdc++.h>
#endif

using namespace std;

#define REP(i,n) for(int i=0;i<(int)(n);++i)
#define RREP(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define FOR(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();++i)
#define LET(x,a) __typeof(a) x(a)
//#define IFOR(i,it,c) for(__typeof((c).begin())it=(c).begin();it!=(c).end();++it,++i)
#define ALL(c) (c).begin(), (c).end()
#define MP make_pair

#define EXIST(e,s) ((s).find(e)!=(s).end())

#define RESET(a) memset((a),0,sizeof(a))
#define SET(a) memset((a),-1,sizeof(a))
#define PB push_back
#define DEC(it,command) __typeof(command) it=command

//debug

#define whole(f,x,...) ([&](decltype((x)) whole) { return (f)(begin(whole), end(whole), ## __VA_ARGS__); })(x)

typedef long long Int;
typedef unsigned long long uInt;
typedef long double rn;

template<class T>
T inf(){
	return numeric_limits<T>::has_infinity?numeric_limits<T>::infinity():(numeric_limits<T>::max()/2);
}

typedef pair<int,int> pii;

#ifdef MY_DEBUG
#include"print.hpp"
#include"debug.hpp"
#endif
// }}}

Int N;
vector<Int> s;

void solve(){
	Int ans = 0;
	for(int S = 1;S <= N - 1;S++){
		int r = (N-1)%S;
		Int sum = 0;
		int left = 0, right = N - 1;
		for(;;){
			right-=S;
			left+=S;
			if(right<0 or left>=N)break;
			if(r==0 and right<=left)break;
			sum += s[left];
			sum += s[right];
			ans = max(ans,sum);
		}
	}
	cout<<ans<<endl;
}

//{{{ main fucnction
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	cin >> N;
    s.assign(N-1-0+1,Int());
    for(int i = 0 ; i < N-1-0+1 ; i++){
        cin >> s[i];
    }
	solve();
	return 0;
}

//}}}