A string ss of length nn can be encrypted by the following algorithm:
- iterate over all divisors of nn in decreasing order (i.e. from nn to 11),
- for each divisor dd, reverse the substring s[1…d]s[1…d] (i.e. the substring which starts at position 11 and ends at position dd).
For example, the above algorithm applied to the string ss="codeforces" leads to the following changes: "codeforces" →→ "secrofedoc" →→ "orcesfedoc" →→ "rocesfedoc" →→ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1d=1).
You are given the encrypted string tt. Your task is to decrypt this string, i.e., to find a string ss such that the above algorithm results in string tt. It can be proven that this string ss always exists and is unique.
Input
The first line of input consists of a single integer nn (1≤n≤1001≤n≤100) — the length of the string tt. The second line of input consists of the string tt. The length of tt is nn, and it consists only of lowercase Latin letters.
Output
Print a string ss such that the above algorithm results in tt.
Examples
Input
10
rocesfedoc
Output
codeforces
Input
16
plmaetwoxesisiht
Output
thisisexampletwo
Input
Copy
1
z
Output
z将某个字符串按下面规则排序：
i从1遍历到n，（1<i<n）,如果n能被i整除，颠倒第1位到第i位的字符
给你加密好的字符串，输出原串
思路：
按照题目所给的例子，找规律，
把加密规则反过来
i从n遍历到1，（1<i<n）,如果n能被i整除，颠倒第1位到第i位的字符
#include <iostream>
#include <cstdio>
#include <deque>
using namespace std;
char s[200];
int str;
int len;
void res(int n)
{
char t;
int i,j;
for(i=0;i<n/2;i++)
{
t=s[i];
s[i]=s[n-i-1];
s[n-i-1]=t;
}
//str+=len/2;
//len/=2;
}
int main()
{
int n,i;
scanf("%d",&n);
getchar();
gets(s);
for(i=2;i<=n;i++)
{
if(n%i==0)
res(i);
}
puts(s);
return 0;
}