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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

var
  x,y:array[0..100000]of char;
  ans:int64;
  i,n,m:longint;
Function gcd(x,y:longint):int64;
begin
  if y=0 then exit(x) else
  exit(gcd(y,x mod y));
end;
begin
  readln(n,m);
  for i:=1 to n do
    read(x[i]);
  readln;
  for i:=1 to m do
    read(y[i]);
  for i:=0 to n-1 do
    if i*m mod n=0 then
      if x[i+1]<>y[i*m div n+1] then
        begin
          write(-1);
          exit;
        end;
  ans:=n*m div gcd(n,m);
  write(ans);
end.