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 Kotlin (1.3.71) (judge verdict: Runtime Error)

import java.util.*

fun main(args: Array<String>) {
  val sc = Scanner(System.`in`)
  val n = sc.nextInt()
  val m = sc.nextInt()
  val s = sc.next()
  val t = sc.next()

  if (s[0] != t[0]) {
    println(-1)
    return
  }

  val l = lcm(n, m)

  val ln = l / n
  val lm = l / m

  val nArray = Array(l) { '0' }
  var ni = 0
  var nn = ni * ln
  while (nn < l) {
    nArray[nn] = s[ni]
    ni++
    nn = ni * ln
  }
  val mArray = Array(l) { '0' }
  var mi = 0
  var mm = mi * lm
  while (mm < l) {
    mArray[mm] = t[mi]
    mi++
    mm = mi * lm
  }


  var ok = true
  for (i in 0 until l) {
    if (nArray[i] == '0') continue
    if (mArray[i] == '0') continue
    
    if (nArray[i] != mArray[i]) {
      ok = false
      break
    }
  }

  if (ok) {
    println(l)
    return
  }
  println(-1)

}

// 最大公約数
private fun gcd(a: Int, b: Int): Int {
  if (b > 0) {
    return gcd(b, a % b)
  }
  return a
}

// 最小公倍数
private fun lcm(a: Int, b: Int): Int {
    return a / gcd(a, b) * b
}