vector容器的reverse函数的使用

题目：

B. Reversing Encryption

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A string $s$

of length $n$

can be encrypted by the following algorithm:

• iterate over all divisors of $n$

in decreasing order (i.e. from $n$ to $1$

), for each divisor $d$, reverse the substring $s[1\dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$

• ).

For example, the above algorithm applied to the string $s$

=" codeforces" leads to the following changes: " codeforces" $\to$ " secrofedoc" $\to$ " orcesfedoc" $\to$ " rocesfedoc" $\to$ " rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$

).

You are given the encrypted string $t$

. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$

always exists and is unique.

Input

The first line of input consists of a single integer $n$

( $1\le n\le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$

, and it consists only of lowercase Latin letters.

Output

Print a string $s$

such that the above algorithm results in $t$

.

Examples

Input

Copy

10
rocesfedoc

Output

Copy

codeforces

Input

Copy

16
plmaetwoxesisiht

Output

Copy

thisisexampletwo

Input

Copy

1
z

Output

Copy

z

Note

The first example is described in the problem statement.

大体意思就是不停地对称，从第一个开始，隔一个，以当前的这个字符为对称轴，两边同时旋转，一直到中间停止。

一开始我是用char做的，打了打之后，发现对称的时候，两边的字符串得到长度必须相等才能对称，比如说 12345，你要以2为对称轴进行对称时，2的左边就很难弄，所以这个之后可以用string，string函数中还带有很多强大的库函数。

代码如下：

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>

using namespace std;
bool cmp(int t1,int t2){
return t1<t2;
}
int main(){
vector<int>wakaka;
int n;
cin>>n;
wakaka.clear();
string s;
cin>>s;
for(int i=1;i<=n;i++){
if(n%i==0){
wakaka.push_back(i);
}
}
sort(wakaka.begin(),wakaka.end(),cmp);
int len=wakaka.size();
for(int i=0;i<len;i++){
int t=wakaka[i];
reverse(s.begin(),s.begin()+t);//字符反转函数
}
cout<<s<<endl;
return 0;
}