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: Runtime Error)

//include
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cmath> 
#include <iomanip>
#include <math.h>
#include <utility>
#include <functional>
#include <cstdlib>
//using
using namespace std;
using vl = vector <long long>;
using vs = vector <string>;
using vc = vector <char>;
using ll= long long;
using vvl = vector<vector<ll> >;
using vvc = vector<vector<char> >;
//vector<vector<char> > hyou(N, vector<char>(N));
using vd = vector <double>;
//define
//#define int long long
#define rep(i,n) for(int i=0; i<n; i++)
#define print(n) cout<<n<<endl;
#define sortp(d) sort(d.begin(),d.end()) //1234
#define sortm(d) sort(d.rbegin(),d.rend()) //4321
//素数判定　O(√A) 
bool is_prime(int x){
 if(x<=1) return false;
 for(int i=2;i*i<=x;i++)
 if(x%i==0) return false;
 return true;
 }
 //各位の和　
int wa(ll x)
{
ll sum = 0;
while(x!=0){
  sum += x% 10;
  x /= 10;
}
return sum;
}
/*順列生成
do {
  // 順列に対する処理
} while (next_permutation(配列変数.begin(), 配列変数.end()));*/

//絶対値　abs()
//文字→数字　'0' アルファベット→数字　'a'　'A'が65番,'a'が97番だわ
//型キャスト　char('a'+1)
//reverse 回文
//if(isupper('A'))->大なら真


//グローバル変数宣言
ll p=0,q=0,r=0;

int main()
{
  string s,t,u;
  cin>>s>>t;

  rep(j,s.size())
  {
    rep(i,s.size())
    {
      if(i==0)u[i]=s[s.size()-1];
      else u[i]=s[i-1];
    }
    rep(i,s.size())
    {
      s[i]=u[i];
    }
    if(s==t)
    {
      print("Yes") return 0;
    }
  }
  print("No")
}