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>
#include <stdlib.h>
#include <math.h>

int euku(int,int);

int main(){
  int i;
  int n,m;
  int p;
  int l=0;

  scanf("%d %d",&n,&m);

  char s[100000];
  char t[100000];

  scanf("%s",s);
  scanf("%s",t);

  if(s[0]!=t[0]){
    l++;
  }
  p = n*m/euku(n,m);
  char x[100000000];

  for(i=0; i<=p-1; i++){
    x[p]='!';
  }

  if(l==0){
    if(p!=n && p!=m){
      l=0;
    }else{
      if(n>=m){
        for(i=0; i<=n-1; i++){
          x[i*p/n+1]=s[i];
        }
        for(i=0; i<=m-1; i++){
          if(x[i*p/m+1]!=t[i] && x[i*p/m+1]!='!'){
            l++;
          }
        }
      }else if(n<m){
        for(i=0; i<=m-1; i++){
          x[i*p/m+1]=t[i];
        }
        for(i=0; i<=n-1; i++){
          if(x[i*p/n+1]!=s[i] && x[i*p/n+1]!='!'){
            l++;
          }
        }
      }
    }
  }
  if(l==0){
    printf("%d\n",p);
  }else if(l>0){
    printf("-1\n");
  }

  return 0;
}

int euku(int a,int b){
  int r;
  if(a>=b){
    r = a % b;
    while(r!=0){
      a = b;
      b = r;
      r = a % b;
    }
    return b;
  }else{
    r = b % a;
    while(r!=0){
      b = a;
      a = r;
      r = b % a;
    }
    return a;
  }
}