999B - Reversing Encryption

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.

题意：给一个长度为n的字符串，经过他的因子个数的几次翻转，让你求最初的字符串。

题解：模拟 那就从2开始判断是不是n的因子，是的话，就翻转从前面到因子的字符串

#include<bits/stdc++.h>
using namespace std;
int b[110],cnt,n;
string s,a;
int main()
{
    cin>>n>>s;
    for(int i=2; i<=n; i++)
        if(n%i==0)
            b[++cnt]=i;
    for(int i=1; i<=cnt; i++)
        for(int j=0; j<b[i]/2; j++)
            swap(s[j],s[b[i]-j-1]);
    cout<<s<<endl;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;char s[110];
    cin>>n>>s;
    for(int i=2;i<=n;i++)
    if(n%i==0)
    reverse(s,s+i);
    cout<<s<<endl;
}

n=int(input())
s=input()
for i in range(2,n+1):
    if n%i==0:
        s=s[i-1::-1]+s[i:]
print(s)