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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
long long lcm(long long u, long long v){
    if(u == 0 || v == 0){
        return 0;
    }else{
        long long s=u*v;
        long long r;
        while (0 != v) {
        r = u % v; u = v; v = r;
        }
        s /= u;
        return s;
    }
}
int gcd(long long u, long long v) {
  long long r;
  while (0 != v) {
    r = u % v; u = v; v = r;
  }
  return u;
}
 
int main(){
    long long n,m;
    string s,t;
    cin >> n >> m;
    cin >> s >> t;
    long long l = lcm(n,m);
    long long o = gcd(n,m);
        long long x = n/o;
        long long y = m/o;
        for(long long i=0;i<o;i++){
            if(s[x*i] != t[y*i]){
                goto dame;
            }
        }
        goto oke;
    
    dame:
    cout << "-1" << endl;
    goto owari;
    oke:
    cout << l << endl;
    owari:
    return 0;
    
}