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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
struct ss{
	char a;
	int b;
}s1[101];
int main(){
	int i,j,c1,t=0,flag;
	char t1[101],t2[101];
	gets(t1);
	gets(t2);
	c1=strlen(t1);
	for(i=0;i<c1;i++)
	{
		s1[i].b=0;
	}
	for(i=0;i<c1;i++){
		flag=0;
		for(j=0;j<t;j++)
			if(s1[j].a==t1[i])
		{
			s1[j].b++;
			flag=1;
			break;
		}
		if(flag==0)
		{s1[t].a=t1[i];
		 s1[t].b=1;
		 t++;
		}
	}
for(i=0;i<c1;i++)
	{
		flag=0;
		for(j=0;j<t;j++)
		{
			if(t2[i]==s1[j].a)
			{
				s1[j].b--;
				flag=1;
				break;
			}
		}
		if(flag==0)
			break;
	}
	if(flag==1)
	for(i=0;i<t;i++)
	if(s1[i].b!=0)
	{	flag=0;
			break;
	}
	if(flag==0)
		printf("No\n");
	else printf("Yes\n");
	return 0;
}