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()
{
    int l,a,b,c,sum=0,temp=0,p=0,er=0,sp=0,o=0;
    char ch[100000];
    
    gets(ch);
    l=strlen(ch);
    int i,ty=0,se=0,re=0;
    if(l<5&&l<11)
    {
        printf("WA\n");
        return 0;
    }
    if(ch[0]=='A')
    {
        sum=1;
    }
    if(ch[2]=='C'||ch[l-2]=='C')
    {
        p=1;
    }
    for(i=0;i<l;i++)
    {
       if(ch[i]=='C'||ch[i]== 'A')
        {
            temp++;
        }
       if(ch[i]>='a'&&ch[i]<='z')
        {
            re++;
            }
        
    }
    for(i=0;i<l;i++)
    {
        if(ch[i]=='C')
        {
            ty++;
        }
    }
    if(ty==1)
    {
        sp=1;
    }
    int st=l-temp;
    if(st==re)
    {
        int se=1;
    }
    if(temp>1)
    {
        er=1;
    }
    o=sum+p+sp;
    if(o==3)
    {
        printf("AC\n");
    }
    else
    {
        printf("WA\n");
    }
}