Problem:
Score : 300 points

Problem Statement

Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.

Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Constraints

1 \leq K \leq 200

K is an integer.

Input

Input is given from Standard Input in the following format:

K

Output

Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.

Sample Input 1

2

Sample Output 1

9

\gcd(1,1,1)+\gcd(1,1,2)+\gcd(1,2,1)+\gcd(1,2,2)
+\gcd(2,1,1)+\gcd(2,1,2)+\gcd(2,2,1)+\gcd(2,2,2)
=1+1+1+1+1+1+1+2=9

Thus, the answer is 9.

Sample Input 2

200

Sample Output 2

10813692

Solution in Pascal (FPC 3.0.4) (judge verdict: Accepted)

program main;
var
	n, sum : integer;
    i, j, k : integer;
    
function gcd(x, y : integer) : integer;
begin
	while true do
    begin
   		x := x mod y;
        if x = 0 then break;
        y := y mod x;
        if y = 0 then break;
    end;
    gcd := x + y;
end;
    
begin
	sum := 0;
	readln(n);
    for i := 1 to n do
    	for j := 1 to n do
        	for k := 1 to n do
            	sum := sum + gcd(gcd(i, j), k);
    writeln(sum);
end.