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
	n,m,t1,i,j:int64;
	s,t:ansistring;
function gcd(x,y:longint):longint;
var
	r,t:longint;
begin
	if x<y then
	begin
		t:=x; x:=y; y:=t;
	end;
        if y=0 then exit(x);
	r:=x mod y;
	while r<>0 do
	begin
		x:=y; y:=r;
		r:=x mod y;
	end;
	exit(y);
end;
begin
	readln(n,m);
	readln(s); readln(t);
	t1:=n*m div gcd(n,m);
	i:=1; j:=1;
	while (i<=n) and (j<=m) do
	begin
		if s[i]<>t[j] then
		begin
			writeln(-1);
			halt;
		end;
		i:=i+t1 div m;
		j:=j+t1 div n;
	end;
	writeln(t1);
end.