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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
char s[100100],t[100100];
long long gcd(long long x,long long y)
{
    if(!y)
        return x;
    return gcd(y,x%y);
}
int main()
{
    long long n,m;
    scanf("%lld%lld",&n,&m);
    scanf("%s",s+1);
    scanf("%s",t+1);
    long long Gcd=gcd(n,m);
    int N=n/Gcd;
    int M=m/Gcd;
    for(int i=1,j=1;i<=n,j<=m;i+=N,j+=M)
        if(s[i]!=t[j])
        {
            puts("-1");
            return 0;
        }
    printf("%lld\n",n*m/Gcd);
    return 0;
}