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 <stdlib.h>
#include <math.h>
#include <string.h>
int euc(int a,int b){
    int tmp;
     if(a<b){
    tmp = a;
    a = b;
    b = tmp;
  }
 int r;
    int x=a*b;
  r = a % b;
  while(r!=0){
    a = b;
    b = r;
    r = a % b;
  }
  x=x/b;
  return x;

}

int main()
{
    int a,b,n,m,k,i=1,j,l;
    long ans=-1;
    char s[100001],t[100001];
    scanf("%d%d",&n,&m);
    scanf("%s",s);
    scanf("%s",t);
    k=euc(n,m);
    if(n>=m){
        l=n;
    }else{
        l=m;
    }
    if(s[0]!=t[0]){
        ans=-1;
    }else{
        a=k/n;
        b=k/m;
        for(j=1;j<l;j++){
        while(1){
            if(i*n>k*j||i*m>k*j){
                break;
            }
            if(s[i*b*j]!=t[i*a*j]){
                break;
            }else{if((i+1)*n>k*j||(i+1)*m>k*j){
                ans=k*j;
            }
                
            }
            i++;
        }
        if(ans!=-1)break;
    }
    }
    if(ans==-1){
        printf("-1");
    }else{
        printf("%ld",ans);
    }
    
    return 0;
}