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>
//ポインタで元のTを操作する
void charRot(char enT[],int length);

int main(void){
    char S[101],T[101];
    scanf("%s %s",S, T);
    //Tの文字列長さ取得
    int Tlength = -1;
    while(T[++Tlength] != '\0');
    //for(Tlength=0;T[Tlength]!='\0';Tlength++);
    //処理
    int i, flag=0;
    char* pt = T;
    for(i=0; i<Tlength; i++){//一致したらフラグ立ててbreak
        if(strcmp(S,T)==0) flag++;
        charRot(pt, Tlength);
        if(flag) break;
    }
    if(flag) printf("Yes\n");
    else printf("No\n");
    return 0;
}
//後ろの文字を一番前に移動させる関数
void charRot(char enT[], int length){
    char temp = enT[length-1];
    int i;
    for(i = length-1; i > 0; i--) enT[i] = enT[i-1];
    enT[0] = temp;
    return;
}