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: Wrong Answer)

#include <bits/stdc++.h>

#define REP(i,n) for(int i=0;i<(int)(n);i++)
#define ALL(x) (x).begin(), (x).end()

using namespace std;

template <typename T> T &chmin(T &a, const T &b) {return a = min(a, b);}
template <typename T> T &chmax(T &a, const T &b) {return a = max(a, b);}

using ll = long long;
using ld = long double;

const int INF = 1e9;
const ld eps = 1e-9, pi = acos(-1.0);

ll gcd(int a, int b){
  if(a<b) swap(a, b);
  if(b==0) return a;
  return gcd(b, a%b);
}

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

int main(){
  ll m, n;
  cin >> n >> m;
  string s, t;
  cin >> s >> t;
  ll x = gcd(n,m);
  bool exist = true;
  for(ll i=0; i<x; i++) if(s[i*(n/x)]!=t[i*(m/x)]) exist = false;
  cout << (exist ? lcm(n,m) : -1) << endl;
  return 0;
}