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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

function toInt (d) {
  return parseInt(d)
}

function Main (input) {
  const data = input.split("\n")
  const fields = data[0].split(' ').map(toInt)
  const N = fields[0]
  const M = fields[1]
  const S = data[1]
  const T = data[2]

  if (S.charAt(0) !== T.charAt(0)) {
    console.log(-1)
    return
  }

  function gcd_two_numbers (x, y) {
    x = Math.abs(x);
    y = Math.abs(y);
    while (y) {
      var t = y;
      y = x % y;
      x = t;
    }
    return x;
  }

  function lcm_two_numbers (x, y) {
    if ((typeof x !== 'number') || (typeof y !== 'number'))
      return false;
    return (!x || !y) ? 0 : Math.abs((x * y) / gcd_two_numbers(x, y));
  }

  const step = lcm_two_numbers(M, N)

  for (var i = step; i <= N * M; i += step) {
    var lcm = lcm_two_numbers(i / N, i / M)
    var f = false
    for (var j = lcm; j < i; j += lcm) {
      if (S.charAt(j / i * N) !== T.charAt(j / i * M)) {
        f = true
        break
      }
    }
    if (f) continue

    console.log(i)
    return
  }

  console.log(-1);
}

Main(require("fs").readFileSync("/dev/stdin", "utf8"))