Problem:
Score : 200 points

Problem Statement

You are given string S and T consisting of lowercase English letters.

Determine if S equals T after rotation.

That is, determine if S equals T after the following operation is performed some number of times:

Operation: Let S = S_1 S_2 ... S_{|S|}. Change S to S_{|S|} S_1 S_2 ... S_{|S|-1}.

Here, |X| denotes the length of the string X.

Constraints

2 \leq |S| \leq 100

|S| = |T|

S and T consist of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S
T

Output

If S equals T after rotation, print Yes; if it does not, print No.

Sample Input 1

kyoto
tokyo

Sample Output 1

Yes

In the first operation, kyoto becomes okyot.

In the second operation, okyot becomes tokyo.

Sample Input 2

abc
arc

Sample Output 2

No

abc does not equal arc after any number of operations.

Sample Input 3

aaaaaaaaaaaaaaab
aaaaaaaaaaaaaaab

Sample Output 3

Yes

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define REP(i, n) for (int i = 0; i < (n); i++)

const long double EPS = 1e-10;
const long long INF = 1e18;
const long double PI = acos(-1.0L);


ll gcd(ll a, ll b){  if(b==0){  return a; }   return gcd(b , a%b); }


int main() {

string s, t;
cin>>s>>t;


unordered_map<char , int> umap1,umap2;


for(int i=0;i<s.length();i++){
  umap1[s[i]]++;
}
for(int i =0;i<t.length();i++){
  umap2[t[i]]++;
}

if(s.length()!=t.length()){
  cout<<"No\n";
  return 0;
}

for(auto i: umap1){
  if(umap1[i.first]!= umap2[i.first]){
    cout<<"No\n";
     return 0;
  }
 
}

cout<<"Yes\n";

  
  return 0;
}