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>
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp> 
#include <functional> // for less 
#define pb push_back
#define For(i,s,e) for (ll i=(s); i<(e); i++)
#define Debug_array(a,n) for (ll i=(0); i<(n); i++) cout<<a[i]<<" "
#define Foe(i,s,e) for (ll i=(s); i<=(e); i++)
#define Fod(i,s,e) for (ll i=(s)-1; i>=(e); i--)
#define Mod 1000000007
#define pii pair<ll,ll>
#define fi first
#define se second
#define endl "\n"
#define mp make_pair
#define big_prime 15486277
#define bigger_prime 179424697
#define biggest_prime 32416188691
//#define random_shuffle(indices.begin(), indices.end());
//std::random_device rd;
//std::mt19937 g(rd());
//std::shuffle(v.begin(), v.end(), g);
using namespace __gnu_pbds; 
using namespace std;

typedef long long ll;
typedef tree<ll, null_type, less<ll>, rb_tree_tag,tree_order_statistics_node_update> ordered_set; 

#define MAX 5002

int main(){
	ios_base::sync_with_stdio(false); 
	// cin.tie(NULL); cout.tie(NULL);
	
	ll n;cin>>n;
	ll a[n];For(i,0,n) cin>>a[i];

	ll ans=0;
	For(C,1,n){
		ll k = 1;
		while(k*C<n-1){
			ll sum = 0;
			ll A = n-1-k*C;
			ll B = A-C;
			if(B<=0 || A<=0 || B>=A || A==2*B) {
				k++;
				continue;
			}
			ll j = 0;
			bool toggle=1;
			while(j<n-1){
				sum+=a[j];
				if(toggle) j+=A;
				else j-=B;
				toggle=!toggle;
			}
			ans = max(ans,sum);
			k++;
		}
	}

	cout<<ans<<endl;








	
}