D. Ice Sculptures

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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.

Sample test(s)
input
81 2 -3 4 -5 5 2 3
output
14
input
61 -2 3 -4 5 -6
output
9
input
61 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.

解题说明：此题意思很明确，求一个正n边形，满足多边形的顶点的数字之和最大。题目上说的很明确，可以用暴力来做，估计除此之外也没有什么好方法。写出来以后提交得到了好几次WA，经过测试发现此题还存在几个隐含的限制，首先顶点的数目必须大于2（否则构成不了图形），其次是总数n必须要是每次迭代步长的整数倍（否则无法封闭），改过后就AC了。

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int main(){int n;int i,j;int step;long max;long temp;int sum;int a[20001];max=0;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);max+=a[i];}for(step=2;step<n;step++){if(n%step==0){for(j=0;j<step;j++){temp=0;sum=0;for(i=j;i<n;i+=step){temp+=a[i];sum++;}if(temp>max&&sum>=3){max=temp;}}}}printf("%ld\n",max);return 0;}