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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = double;
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define all(v) (v).begin(),(v).end()
#define siz(v) (ll)(v).size()
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define repi(i,x,n) for(ll i=x;i<(ll)(n);i++)
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
const ll mod = 1000000007;
const ll INF = 1000000099;
vector<ll> dx={-1,1,0,0},dy={0,0,-1,1};
//cin.tie(0);
//ios::sync_with_stdio(false);


signed main(){
    ll n,ans=0;cin>>n;
    vector<ll> v(n);
    rep(i,n)cin>>v[i];
    ans=0;

    for(int i=1;i < n-1;i++)//a-b
    {
        ll res=0;
        set<ll> se;
        for(int j=1;i*(j+1)<n-1;j++)//係数
        {
            if(!se.count(i*j) && !se.count(n-1-i*j) && i*j!=n-1-i*j){
                res+=v[i*j]+v[n-1-i*j];
                ans=max(ans,res);
                se.emplace(i*j);
                se.emplace(n-1-i*j);
                //cout<<i*j<<" "<<n-1-i*j<<endl;
            }
        }
        se.clear();
    }
    cout<<ans<<endl;
}