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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main{
    public static void main (String[] args){
        Scanner scan = new Scanner(System.in);
//
        String str1 = scan.nextLine();
        String str2 = scan.nextLine();
        if(str1.length() != str2.length()){
            System.out.println("error");
            return;
        }
        if(str1.length() == 1){
            if(str1.equals(str2)){
                System.out.println("Yes");
            }else{
                System.out.println("No");
            }            
            return;
        }
//
        String[] strArry1 = new String[str1.length()];
        String[] strArry2 = new String[str2.length()];
//
        setArray(strArry1, str1);
        setArray(strArry2, str2);
//
        String[] wstrArry = new String[str1.length()];
            for (int i = 0; i < strArry1.length; i++){
                cnvArry(wstrArry,strArry1);
                if(comp(wstrArry,strArry2) == 0){
                    System.out.println("Yes");
                    return;
                }
            }            
            System.out.println("No");
            return;
    }
/*
    split strings
*/
    public static void setArray(String[] array, String str){
        for (int i = 0; i < array.length; i++){
            array[i] = String.valueOf(str.charAt(i));
        }
    }
/*
    compare strings
*/
    public static int comp(String[] wstrArry, String[] strArry2){
        for (int i = 0; i < wstrArry.length; i++){
            if(wstrArry[i].equals(strArry2[i])){
                continue;
            }else{                
                return 1;
            }
        }
        return 0;        
    }
/*
    convert
*/
    public static void cnvArry(String[] wstrArry, String[] strArry1){
        wstrArry[0] = strArry1[strArry1.length - 1];
        System.out.println(wstrArry[0]);
        for (int j = 0; j < strArry1.length-1; j++){
            wstrArry[j+1] = strArry1[j];
        }
//
        for (int k = 0; k < strArry1.length; k++){
            strArry1[k] = wstrArry[k];
        }
    }
}