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: Wrong Answer)

def getdup(a, b)
  res = {}
  a.each do |key, val|
    b.each do |k1, v1|
      res[key] = val if val == v1
    end
  end
  res
end

N, M = gets.chomp.split().map(&:to_i)
S = gets.chomp
T = gets.chomp
n = N.lcm(M)
i = 1
loop do
  l = n*i
  if l > 100000 || S[0] != T[0]
    puts -1
    break
  else
    valN = {}
    p = 0
    loop do
      break if p > (N-1)
      if p == 0
        valN[p] = 1
      else
        valN[p] = p*(l/N)+1
      end
      p += 1
    end
    valM = {}
    p = 0
    loop do
      break if p > (M-1)
      if p == 0
        valM[p] = 1
      else
        valM[p] = p*(l/M)+1
      end
      p += 1
    end
    a = getdup(valN, valM)
    b = getdup(valM, valN)
    status = false
    a.keys.length.times do |s|
      if S[a.keys[s]] != T[b.keys[s]]
        status = true
      end
    end
    unless status
      puts l
      break
    end
  end
  i += 1
end