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)

local function gcd(a,b)
    if b>0 then
        return gcd(b,a%b)
    else
        return a
    end
end

local n,m=io.read("*n","*n","*l")
local s=io.read()
local t=io.read()

local l=n*m//gcd(n,m)
local ln=m//gcd(n,m)
local lm=n//gcd(n,m)

local x={}
for i=1,n do
    x[(i-1)*ln+1]=s:sub(i,i)
end

for i=1,m do
    if not x[(i-1)*lm+1] then
        x[(i-1)*lm+1]=t:sub(i,i)
    elseif x[(i-1)*lm+1]~=s:sub(i,i) then
        print(-1)
        return
    end
end
print(l)