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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(int i=0; i<(n); i++)
#define REP2(i,x,n) for(int i=x; i<(n); i++)
#define ALL(n) begin(n),end(n)
struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;
const long long INF = numeric_limits<long long>::max();

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

int main() {
  int N, M;
  cin >> N >> M;
  string S, T;
  cin >> S >> T;
  int64_t g = gcd(N, M);
  int64_t L = N * M / g;
  // cout << L << endl;
  if (g == 1) {
    if (S.at(0) == T.at(0)) cout << L << endl;
    else cout << -1 << endl;
  }
  else {
  bool OK = true;
  for (int i = 0; i < L; i += L / g) {
    // cout << i % (L / N) << ' ' << i % (L / M) << endl;
    if (i % (L / N) == 0 && i % (L / M) == 0) {
      // cout << S.at(i / (L / N)) << ' ' << T.at(i / (L / M)) << endl;
      if (S.at(i / (L / N)) != T.at(i / (L / M))) OK = false;
    }
  }
  if (OK) {
    cout << L << endl;
  }
  else cout << -1 << endl;
  }
  return 0;
}