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 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int abs(int a,int b){
  if(a < b){
    int temp = a;
    a = b;
    b = temp;
  }
  if(b == 0){
    return a;
  }
  
  abs(b,a%b);
}

int main(void){
  int n , m;
  scanf("%d %d",&n ,&m);

  char s[n] , t[m];
  scanf("%s %s",s ,t);

  int gcd = abs(n,m);
  int gn = n/gcd , gm = m/gcd;


  int flag = 0;
  for(int i = 0;i < gcd;i ++){
    for(int j = 0;j < gcd;j ++){
      if(gn*i == gm*j){
        //printf("%c %c\n",s[i],t[j]);
        if(s[i] != t[j]){
          flag = 1;
          i += n;
          j += m;
        }
      }
    }
  }

  if(flag == 0){
    printf("%lld\n",n/gcd*m);
  }

  else{
    printf("-1\n");
  }

  return 0;
}