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>

void my_push_front(char *str, char c)
{
    if( c == '\0' ) return;
    const int len = strlen(str);
    for(int i = len; i >= 0; --i) {
        str[i+1] = str[i]; //  文字列を1文字後ろにずらす
    }
    *str = c;      // 先頭に1文字挿入
}

int main(void){
	char s[101];
	char t[101];
	char tmp;
	int len;
	scanf("%s",s);
	scanf("%s",t);
	len = strlen(s);
	int flag= 0;
	int i;
	if( strcmp(s,t)==0){
		flag =1;
	}
	for(int j=0;j<len;j++){
		for(i=len;i>=0;i--){
			s[i+1] = s[i];//これを行うと文字列が右に一つずれる。
		}
		s[0] = s[len];
		s[len] ='\0';
		if( strcmp(s,t)==0){
		flag =1;
		}
	}
	if( flag == 1){
		printf("Yes");
	}else{
		printf("No");
	}
	return 0;
}