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 <string.h>

int gcd(int a, int b) {
    if (b == 0)
        return a;

    return gcd(b, a % b);
}

int lcm(int a, int b) {
    return a * b / gcd(a, b);
}



int main(){
  int N,M;
  scanf("%d%d",&N,&M);
  char S[N];
  char T[M];
  scanf("%s",S);
  scanf("%s",T);
  int i;
  int flag=0;

  int ans=-1;
  if(S[0]!=T[0]){
    flag=1;
  }
  if(N==M){
    for(i=0;i<N;i++){
      if(S[i]!=T[i]){
	flag=1;
      }
    }
  }else{
    for(i=0;i<N;i++){
      if(i*M%N==0){
	if(S[i]!=T[i*M/N]){
	  flag=1;
	}
      }
    }
  }
  if(flag==0){
    ans=lcm(N,M);
  }
  printf("%d",ans);
  return 0;
}