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>


int main()
{
    char str1[100],str2[100],temp;
    int i,c=0,l,l2,d=0,j;
    scanf("%s%s",&str1,&str2);
    l=strlen(str1);
    l2=strlen(str2);
    for(i=0;i<l;i++){
        if((str1[i]>='a' && str1[i]<='z') && (str2[i]>='a' && str2[i]<='z')){
            c++;
        }
    }
    if((l>=2 && l<=100) && (l2>=2 && l2<=100) && (l==l2) && (l==c) ){

        for(j=0;j<l;j++){
            temp=str1[l-1];
            for(i=1;i<l;i++){
            str1[l-i]=str1[l-i-1];
            }
            str1[0]=temp;
            if(!strcmp(str1,str2)){
                printf("YES\n");
                break;
            }
            else{
                d++;
            }
        }
        if(d==l){
            printf("NO\n");
        }

    }
}