Problem:
Score : 300 points

Problem Statement

You are given a string S of length N and another string T of length M.
These strings consist of lowercase English letters.

A string X is called a good string when the following conditions are all met:

Let L be the length of X. L is divisible by both N and M.

Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.

Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.

Determine if there exists a good string. If it exists, find the length of the shortest such string.

Constraints

1 \leq N,M \leq 10^5

S and T consist of lowercase English letters.

|S|=N

|T|=M

Input

Input is given from Standard Input in the following format:

N M
S
T

Output

If a good string does not exist, print -1; if it exists, print the length of the shortest such string.

Sample Input 1

3 2
acp
ae

Sample Output 1

6

For example, the string accept is a good string.
There is no good string shorter than this, so the answer is 6.

Sample Input 2

6 3
abcdef
abc

Sample Output 2

-1

Sample Input 3

15 9
dnsusrayukuaiia
dujrunuma

Sample Output 3

45

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
long long llmax(long long a, long long b){return a>b?a:b;}
long long llmin(long long a, long long b){return a<b?a:b;}
long long llgcd(long long a, long long b){return llmin(a,b)==0?llmax(a,b):llgcd(llmin(a,b),llmax(a,b)%llmin(a,b));}
int main(void){
    long long n,m;
    scanf("%lld%lld",&n,&m);
    char s[n+1],t[m+1];
    scanf("%s%s",s,t);
    for (long long i=0; i<llgcd(n,m); i++){
        if (s[i*n*m/llgcd(n,m)/m]!=t[i*n*m/llgcd(n,m)/n]){
            printf("-1\n");
            return 0;
        }
    }
    printf("%lld\n",n*m/llgcd(n,m));
	return 0 ;
}