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: Accepted)

#include<stdio.h>
#include<string.h>
int def(char X[],char Y[]);
int main()
{
    char A[10000],B[10000];
    scanf("%s %s",A,B);
    int m,n,ans=0;
    m=strlen(A);
    n=strlen(B);
    if(m!=n)
    {
        printf("No\n");
        return 0;
    }
    ans=def(A,B);
    if(ans==1)
    {
        printf("Yes\n");
        return 0;
    }
    if(ans==-1)
    {
        printf("No\n");
        return 0;
    }
    return 0;
}

int def(char X[],char Y[])
{
    int M,i=0,j=0,put=0,K=0;
    char C[100];
    M=strlen(X);

    while(K<M)
    {
j=0;

    for(i=0;i<M;i++)
    {
        if(X[i]==Y[i])
        {
            j++;
        }
    }
    if(j==M)
    {
        put=1;
        return put;
    }

    if(put==1)
    break;
C[0]=X[M-1];
    for(i=M-1;i>0;i--)
    {

        X[i]=X[i-1];
    }
    X[0]=C[0];


    if(put==0 && K==M-1)
    {
        put=-1;
        return put;
    }
    K++;
    }


}