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>
#include <math.h>
#include <stdbool.h>
#include <string.h>

int GCD(int x, int y)
{
  int z;
  if(x<y)
  {
    z = x;x = y;y = z;
  }
  z = x%y;
  if(z==0)
    return y;
  else
    return GCD(y, z);
}

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