codeforces158D

Ice Sculptures

CodeForces - 158D

The Berland University is preparing to celebrate the 256-th anniversary of its founding! A specially appointed Vice Rector for the celebration prepares to decorate the campus. In the center of the campus n ice sculptures were erected. The sculptures are arranged in a circle at equal distances from each other, so they form a regular n-gon. They are numbered in clockwise order with numbers from 1 to n.

The site of the University has already conducted a voting that estimated each sculpture's characteristic of t_i — the degree of the sculpture's attractiveness. The values of t_i can be positive, negative or zero.

When the university rector came to evaluate the work, he said that this might be not the perfect arrangement. He suggested to melt some of the sculptures so that:

• the remaining sculptures form a regular polygon (the number of vertices should be between 3 and n),
• the sum of the t_i values of the remaining sculptures is maximized.

Help the Vice Rector to analyze the criticism — find the maximum value of t_i sum which can be obtained in this way. It is allowed not to melt any sculptures at all. The sculptures can not be moved.

Input

The first input line contains an integer n (3 ≤ n ≤ 20000) — the initial number of sculptures. The second line contains a sequence of integers t₁, t₂, ..., t_n, t_i — the degree of the i-th sculpture's attractiveness ( - 1000 ≤ t_i ≤ 1000). The numbers on the line are separated by spaces.

Output

Print the required maximum sum of the sculptures' attractiveness.

Examples

Input

8
1 2 -3 4 -5 5 2 3

Output

14

Input

6
1 -2 3 -4 5 -6

Output

9

Input

6
1 2 3 4 5 6

Output

21

Note

In the first sample it is best to leave every second sculpture, that is, leave sculptures with attractivenesses: 2, 4, 5 и 3.

sol：想了半天都不会，然后看题解，然后惊呼这TM能过？？

for(i=;i*<=n;i++)
    {
        if(n%i==)
        {
            for(j=;j<=i;j++)
            {
                int Sum=;
                for(k=j;k<=n;k+=i)
                {
                    Sum+=a[k];
                }
                ans=max(ans,Sum);
            }
        }
    }

这也太暴力了吧，我怎么算复杂度都是n²及以上啊，为什么能过啊qaq

Ps：不管了，弃了弃了（跪求更好的做法或证明一下复杂度）

4.3updata：这是调和级数的复杂度，似乎是n^1.5。实锤不会算复杂度qaq

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=;
    bool f=;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<)+(s<<)+(ch^); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<)
    {
        putchar('-'); x=-x;
    }
    if(x<)
    {
        putchar(x+''); return;
    }
    write(x/);
    putchar((x%)+'');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=;
int n,a[N];
int main()
{
    int i,j,k,ans=-0x3f3f3f3f;
    R(n);
    for(i=;i<=n;i++) R(a[i]);
    for(i=;i*<=n;i++)
    {
        if(n%i==)
        {
            for(j=;j<=i;j++)
            {
                int Sum=;
                for(k=j;k<=n;k+=i)
                {
                    Sum+=a[k];
                }
                ans=max(ans,Sum);
            }
        }
    }
    Wl(ans);
    return ;
}
/*
input
8
1 2 -3 4 -5 5 2 3
output
14

input
6
1 -2 3 -4 5 -6
output
9

input
6
1 2 3 4 5 6
output
21

input
4
1 -10 1 -10
output
-18
*/