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: Runtime Error)

#include<stdio.h>
#include <string.h>
int main(){
  int N,M,L=0,i,j,k;
  char S[256],T[256],x[256];
  scanf("%d %d",&N,&M);
  
  scanf("%s",S);
  scanf("%s",T);
  L=N;
  
  for(k=0;k=100000;k++){
    if((L%N)==0&&(L%M)==0){
      for(i=0;i<N;i++){
          x[(L/N)*i]=S[i];
      }
	  for(j=0;j<M;j++){
	    if(x[(L/M)*j]='\0' || x[(L/M)*j]==T[j]){
	    	x[(L/M)*j]=T[j];
	    }
	  }
	  if((L/N)*i <= (L/M)*j){
	    	printf("%d",(L/M)*j);
	    	break;
	  }else if((L/N)*i > (L/M)*j){
	  	printf("%d",(L/M)*i);
	  	break;
	  }
    }
  	L+=1;
  }
  if(k==100000)printf("-1");
  return 0;
}