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>
int main(void){
   int a,b,c,d,i,x,tmp,r;
   char a1[100000],b1[100000],a2[100000],b2[100000];
    
    scanf("%d%d",&a,&b);
   gets(a1);
   gets(b1);

    c=a;d=b;
  x=a*b;
  if(a<b){
    tmp=a;
    a=b;
    b=tmp;
  }
  r=a%b;
  while(r!=0){
    a=b;
    b=r;
    r=a%b;
  }
  x=x/b;
  
  for(i=1;i<=c;i++){
      a2[(i-1)*(x/c)+1]=a1[i-1];
  }
  for(i=1;i<d;i++){
      b2[(i-1)*(x/d)+1]=b1[i-1];
  }
  
for(i=1;i<x;i++){
    if(a2[i]!=b2[i]){
        if((a2[i]!='\0')&&(b2[i]!='\0')){
            printf("-1");
            return 0;
        }
        
    }
}
printf("%d",x);
return 0;
}