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.Arrays;
import java.util.Scanner;
public class Main { //B

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String S = sc.next();
		String T = sc.next();
		int l = S.length();

		char[] SS = new char[l*2];
		char[] TT = new char[l*2];

		for(int i=0; i<l; i++) {
			SS[i] = S.charAt(i);
			SS[l+i] = S.charAt(i);
			TT[i] = S.charAt(i);
			TT[l+i] = T.charAt(i);
		}

		int[] judge = new int[l];
		for(int j=0; j<l; j++) {
			int judge1 = 0;
			int judge2 = 0;
			for(int k=0; k<l; k++) {
				//int judge2 = 0;
				if(SS[k] == TT[k+j]) {
					judge2 += 1;
				}
			}if(judge2 == l) {
				judge[j] = 1;
			}
		}Arrays.sort(judge);

		//System.out.println(judge1);

		if(judge[l-1] == 0) {
			System.out.println("No");
		}else {
			System.out.println("Yes");
		}


	}

}