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)

var k,ans,a,b,c,d :integer;

function gcd(p,q:integer):integer;
begin
  if (p mod q =0)then gcd:=q
  else gcd(q,p mod q)
end;

begin
  readln(k);
  ans:=0;
  for a:=1 to k do
  begin
    for b:=1 to k do
    begin
      d:=gcd(a,b);
      for c:=1 to k do
      begin
        inc(ans,gcd(d,c));
      end;
    end;
  end;
  writeln(ans);
end.