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 <cstdio>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <vector>
#include <iostream>
#include <ctime>
#include <unistd.h>

using namespace std;

#define TRACE(x) cerr << #x <<  " = " << x << endl
#define _ << " " <<
#define FOR(i, a, b) for (int i=(a); i<(b); i++)
#define REP(i, n) FOR(i, 0, n)

#define X first
#define Y second

typedef pair<int, int> P;
typedef long long ll;

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

const int MAX = 1<<19;

char s1[MAX], s2[MAX];
int n, m, g;

int main()
{
  scanf("%d%d %s %s", &n, &m, s1, s2);

  ll L = lcm(n, m);
  g = __gcd(n, m);
  int mp = m / g, np = n / g;

  int ok=1;
  for (int a=0; ; a++) {
    ll kn = (ll) np * a;
    ll km = (ll) mp * a;
    if (kn >= n || km >= m) break;
    if (s1[kn] != s2[km])
      ok = 0;
  }

  if (!ok) printf("-1\n");
  else printf("%lld\n", L);

  return 0;
}