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 <iostream>
#include <cassert>
#include <climits>
#include <bitset>
#include <stack>
#include <queue>
#include <iomanip>
#include <limits>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string.h>
#include <tuple>
#include <time.h>
#include <random>

using namespace std;

typedef long long ll;
typedef pair<ll,ll> P;

long long int INF = 3e18;
double Pi = 3.1415926535897932384626;

//vector<ll> G[500005];
//vector<P> tree[500010];
priority_queue <ll> pql;
priority_queue <P> pqp;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop

int dx[8]={1,0,-1,0,1,1,-1,-1};
int dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "DRUL";
//ll bit[500005];
//↓,→,↑,←


#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<(x)<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define re(i,n) for(i=0;i<n;i++);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define deba(x) cout<< #x << " = " << x <<endl


ll mod = 1000000007;
ll rui(ll number1,ll number2){

    if(number2 == 0){
        return 1;
    }else{
        ll number3 = rui(number1,number2 / 2);
        number3 *= number3;
        number3 %= mod;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod;
        }
        return number3;
    }
}
ll gcd(ll number1,ll number2){

    if(number1 > number2){
        swap(number1,number2);
    }
    if(number1 == 0 || number1 == number2){
        return number2;
    }else{
        return gcd(number2 % number1,number1);
    }
}
void YES(bool condition){

    if(condition){
        p("YES");
    }else{
        p("NO");
    }
    return;
}
void Yes(bool condition){

    if(condition){
        p("Yes");
    }else{
        p("No");
    }
    return;
}

/*
void c3_init(){
    fact[0] = rfact[0] = 1;
    for(ll i=1; i<=800000; i++){
        fact[i] = (fact[i-1]*i) % mod;
    }
	rfact[800000] = rui(fact[800000],mod - 2);
	for(ll i=800000; i>=1; i--){
	   rfact[i-1] = rfact[i] * i;
	   rfact[i-1] %= mod;
	}
	return;}
ll c3(ll n,ll r){
    return (((fact[n] * rfact[r]) % mod) * rfact[n-r]) % mod;}
*/

ll n,m,num,sum,ans,a,b,c,d,e,g,h,w,i,j,k,q,l,r;
ll x[500005],y[500005],z[500005];
ll t[500005],table[700005];
//bool used[500005];
bool flag,dame;

int main(){
	cin >> n;
	ll ab;
	for(i=0;i<n;i++){
		cin >> x[i];
	}
	ll num = 0;
	for(i=1;i<=n;i++){
		a = i;
		for(j=1;j*j<=n-1-a;j++){
			if(i == 3 && j == 1)p("!");
			if((n-1-a)%j != 0)continue;
			b = i - j;
			ab = j;
			if(a % ab == 0 && a >= ab && n >= 2 * a)continue;
			if(b <= 0)continue;
			//pe(a);pe(b);
			ans = x[a];
			for(k=a+ab;k<=n;k+=ab){
				ans += x[k];
			}
			ans += x[n-a-1];
			for(k=n-a-ab-1;k>=0;k-=ab){
				ans += x[k];
			}
			//p(ans);
			num = max(num,ans);
			b = i - (n-1-a)/j;
			ab = (n-1-a)/j;
			if(a % ab == 0 && a >= ab)continue;
			if(b <= 0)continue;
			//pe(a);pe(b);
			ans = x[a];
			for(k=a+ab;k<=n;k+=ab){
				ans += x[k];
			}
			ans += x[n-a];
			for(k=n-a-ab;k>=0;k-=ab){
				ans += x[k];
			}
			//p(ans);
			num = max(num,ans);
		}
	}
	p(num);
	return 0;
}