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 Common Lisp (SBCL 1.1.14) (judge verdict: Wrong Answer)

;; Two Abbreviations

(defun f1 (n m s1 s2)
  (let ((k (gcd n m))
        )
    (labels ((inner (p q)
               (if (or (<= n p) (<= m q))
                   t
                   (if (eq (nth p s1)
                           (nth q s2))
                       (inner (+ (+ 1 p) (/ n k))
                              (+ (+ 1 q) (/ m k)))
                       nil))))
      (inner (/ n k) (/ m k)))    
    ))

(defun main ()
  (let ((n (read)) (m (read)))
    (if (f1 n m
            (loop for k from 1 upto n collect (read-char))
            (loop for k from 1 upto m collect (read-char)))
        (princ (lcm n m))
        (princ -1))))
(main)