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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

n, m = gets.chop.split.map(&:to_i)
s = gets.chop.split('')
t = gets.chop.split('')

if s[0] == t[0]

  l = n.lcm(m)

  s_diff = l / n
  t_diff = l / m

  s_arr = []
  t_arr = []

  n.times do |i|
    s_arr << i * s_diff + 1
  end

  m.times do |i|
    t_arr << i * t_diff + 1
  end

  s_is = [0]
  t_is = [0]
  i = 1
  loop do
    next_num = i * t_diff
    if next_num >= n
      break
    else
      s_is << next_num
    end
    i += 1
    # s_is << i * t_diff
    # i += 1
    # if s_is.last >= n
    #   s_is.pop
    #   break
    # end
  end
  i = 1
  loop do
    next_num = i * s_diff
    if next_num >= m
      break
    else
      t_is << next_num
    end
    i += 1
    # t_is << i * s_diff
    # i += 1
    # if t_is.last >= m
    #   t_is.pop
    #   break
    # end
  end

  # p s_is, t_is

  flag = true
  s_is.each do |s_i|
    break unless flag
    t_is.each do |t_i|
      next unless s_arr[s_i] == t_arr[t_i]
      if s[s_i] != t[t_i]
        flag = false
        break
      end
    end
  end

  if flag
    puts l
  else
    puts '-1'
  end

else
  puts '-1'
end