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: Runtime Error)

# include <bits/stdc++.h>
using namespace std;
int gcd(int a,int b)
   {
   	 if (b==0)
   	   return a;
   	 else
   	   return gcd(b,a%b);
   }
int main ()
{
  int M,N;
  string S,T;
  cin>>M;
  cin>>N;
  cin>>S>>T; 
  long long c=M*N/gcd(M,N);
  if (S[0]!=T[0])
    {
    	printf ("%d",-1);
    	exit(-1);
	}
	int flag=1;
     	for (long long x=1;x<M;x++)
     	   {
     	   	 for (long long y=1;y<N;++y)  
     	   	    
					   if ((N*x==M*y)&&(S[x]!=T[y]))
					     {
					     	flag=0;
					     	break;
					     }
			 if (flag==0)
			   break;
			}
	if (flag)
	  printf ("%d\n",c);
	else
	  printf ("%d\n",-1);
	  return 0;
}