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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include"stdio.h"
#include"string.h"
int main()
{
	int ls1,ls2,q=0;
	char s1[105],s2[105],s3[105],s4[105];
	scanf("%s%s",s1,s2);
	ls1=strlen(s1);
	ls2=strlen(s2);
	memset(s3,0,sizeof(s3));
	memset(s4,0,sizeof(s4));
	for(int i=0,h=0;i<25;i++)
		{
			for(int j=0;j<ls1;j++)
			{
				if(s1[j]=='a'+i)
				{
					s3[h]=s1[j];
					h++;
				}
			}	
		}
	for(int i=0,h=0;i<25;i++)
		{
			for(int j=0;j<ls2;j++)
			{
				if(s2[j]=='a'+i)
				{
					s4[h]=s2[j];
					h++;
				}
			}	
		}
	for(int i=0;i<ls1;i++)
	{
		if(s3[i]!=s4[i])
			q=1;
			
	}
	if(q==1)
	printf("No\n");
	else		
	printf("Yes\n");		
}