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 Lua (5.3.2) (judge verdict: Wrong Answer)

function gcd(u, v)
  while v ~= 0 do
    u, v = v, u%v
  end
  return u
end

function lcm(u,v)
  return u * v / gcd(u, v)
end

N, M = string.match(io.read(), "(%d) (%d)")

L = lcm(N,M)
ans = L

oS = L/N
oT = L/M

X = {}
p = 1
for v = 1, N do
  X[p] = io.read(1)
  p = p + oS
end
io.read()

p = 1
for v = 1, M do
  c = io.read(1)
  if not (X[p] == nil or X[p] == c) then
    ans = -1
    break
  end
  p = p + oT
end
io.read()

print(string.format("%d",ans))