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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>


long long int calc_LCM(int ,int);

int main(void){
	
	int N,M,i,j;
	char *S,*T;
	long long int L;	
  
	scanf("%d %d",&N,&M);
	S = (char *)malloc((N+1)*sizeof(char));
	T = (char *)malloc((M+1)*sizeof(char));
	scanf("%s",S);
	scanf("%s",T);
	
	if(S[0] != T[0]){
		printf("-1");
		free(S);
		free(T);
		return 0;
	}
	
	L = calc_LCM(N,M);
  	for(i=0;i<N;i++){
    	for(j=0;j<M;j++){
        	if(i*L/N == j*L/M){
            	if(S[i] != T[j]){
        	      printf("-1\n");
        	   	  free(S);
				  free(T);
	
				  return 0;                
                }
              	else{
                	break;
                }
            }
        }
    }

	printf("%lld",L);

	free(S);
	free(T);
	
	return 0;
}


long long int calc_LCM(int a,int b){
	
	int tmp;
  	long long int x = a*b;
	
	if(a<b){
		tmp = a;
		a = b;
		b = tmp;
	}
	
	tmp = a% b;
	
	while(tmp != 0){
		a = b;
		b = tmp;
		tmp = a % b;
	}
	return x / b;
	
}