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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <algorithm>
#include <bitset>
#define range(i, r) for(int i=0;i<r;i++)
#define ranges(i, l, r) for(int i=l;i<r;i++)
#define vv(a, b, c, d) vector<vector<d> >(a, vector<d>(b, c))
#define vvi std::vector<std::vector<int> >
#define vvl std::vector<std::vector<ll> >
#define MODs 1000000007;
#define MODn 1000000009;
typedef long long int ll;
using namespace std;

ll gcd(ll n, ll m){
    ll maxn = max(n,m);
    ll minn = min(n,m);
    ll r = 1;
    while(r!=0){
      r = maxn % minn;
      maxn = minn;
      minn = r;
    }
    return maxn;
}
int main(int argc, char const *argv[]) {
  ll N, M;
  std::cin >> N >> M;
  string S, T;
  std::cin >> S;
  std::cin >> T;
  ll lcm = (N * M)/gcd(N, M);
  int s=0, t=0;
  bool flag = 1;
  while(true){
    if(s>=N||t>=M) break;
    if(S[s] != T[t]) flag = false;
    s+=lcm/M;
    t+=lcm/N;
  }
  if(flag) std::cout << lcm << '\n';
  else std::cout << -1 << '\n';
  return 0;
}