循环移位(Cycle)

循环移位(Cycle)

---

Description

Cycle shifting refers to following operation on the sting. Moving first letter to the end and keeping rest part of the string. For example, apply cycle shifting on ABCD will generate BCDA. Given any two strings, to judge if arbitrary times of cycle shifting on one string can generate the other one.

Input

There m lines in the input, while each one consists of two strings separated by space. Each string only contains uppercase letter 'A'~'Z'.

Output

For each line in input, output YES in case one string can be transformed into the other by cycle shifting, otherwise output NO.

Example

Input

AACD CDAA
ABCDEFG EFGABCD
ABCD ACBD
ABCDEFEG ABCDEE

Output

YES
YES
NO
NO

Restrictions

0 <= m <= 5000

1 <= |S1|, |S2| <= 10^5

Time: 2 sec

Memory: 256 MB

1. 原理与要点：读入两个字符串A和B，如果两字符串长度不同，则直接输出No。如果长度相同，则把A串接在A串后面形成一个新串，然后查询B是否是新串的子串，如果是则Yes，否则No
2. 遇到的问题：无
3. 时间和空间复杂度：时间复杂度(O(N))，空间复杂度(O(N))。

#include "cstdio"
#include "cstring"

using namespace std;
const int maxn = 2e5 + 100;

char s[maxn], t[maxn];
const int SZ = 1<<20;  //快速io
struct fastio{
    char inbuf[SZ];
    char outbuf[SZ];
    fastio(){
        setvbuf(stdin,inbuf,_IOFBF,SZ);
        setvbuf(stdout,outbuf,_IOFBF,SZ);
    }
}io;
int main() {
    while (scanf("%s %s", s, t) != EOF) {
        if (strlen(s) != strlen(t)) {
            printf("NO
");
            continue;
        }
        int len = strlen(s);
        for (int i = len; i < len * 2; i++) {
            s[i] = s[i - len];
        }
        s[len * 2] = '';
        if (strstr(s, t) != NULL)
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}