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

int gcd(int,int);
int lcm(int,int);
int min(int,int);
int max(int,int);

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

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

  unsigned char s[1000000];
  unsigned char t[1000000];

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

  if(s[0]!=t[0]){
    l++;
  }
  q = gcd(n,m);
  p = lcm(n,m);

    for(i=0; i<=q; i++){
      if(s[i*n/q]!=t[i*m/q]){
        l++;
      }
    }

  if(l==0){
    printf("%lld\n",p);
  }else if(l>0){
    printf("-1\n");
  }

  return 0;
}

int min(int m, int n){
  if(m >= n){
    return n;
  }else {
    return m;
  }
}

int max(int m, int n){
  if(m >= n){
    return m;
  }else {
    return n;
  }
}

int gcd(int m,int n){
  int i;
  int x;
  int a;
  x = min(m,n);
  for(i=1; i<=x; i++){
    if(m%i==0 && n%i==0){
      a = i;
    }
  }
  return a;
}

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