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.split(" ").map(&:to_i)
s = gets.chomp.to_s
t = gets.chomp.to_s

lcm = s.size.lcm(t.size)

x = []

lcm.times do
  x.push("")
end

lcm_s = lcm / s.size

s.size.times do |num|
  x[(1 + lcm_s * num)] = s[num]
end

lcm_t = lcm / t.size

t.size.times do |num|
  if x[(1 + lcm_t * num)] != "" && x[(1 + lcm_t * num)] != t[num]
    puts "-1"
    exit
  end
end

puts lcm