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: Wrong Answer)

#include <iostream>
#include<cstdio>

using namespace std;

int gys(int n,int m)
{
    if(n<m)
    {
        int t=n;
        n=m;
        m=t;
    }
    int a=n%m;
    while(a!=0)
    {
        n=m;
        m=a;
        a=n%m;
    }
    return m;
}

int main()
{
    int n,m;
    char S[100005],T[100005];
    while(scanf("%d %d",&n,&m)==2)
    {
        scanf("%s",S);
        scanf("%s",T);
        int G=gys(n,m);
        int X=n*m/G;
        if(S[0]!=T[0])
        {
            cout<<-1<<endl;
            continue;
        }
        else
        {
            int a=n/G;
            int b=m/G;
            int flag=1;
            for(int i=1; i<=G; i++)
            {
                if(S[i*a]!=T[i*b])
                {
                    flag=0;
                    break;
                }
            }
            if(flag)
            {
                cout<<X<<endl;
            }
            else
            {
                cout<<-1<<endl;
            }
        }




    }
    return 0;
}