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
    i,j,k : integer;
    n : integer;
    sum : longint;
    dp : array [1..200,1..200] of integer;
function gcd(a,b : integer):integer;
var cur,wth:integer;
begin
    wth:=1;
    for cur:=2 to 200 do begin
        while ((a mod cur) = 0) and ((b mod cur) = 0) do begin
            a:=(a div cur);
            b:=(b div cur);
            wth:=wth*cur;
        end;
    end;
    gcd:=wth;
end;
begin
    for i:=1 to 200 do
        for j:=1 to 200 do
            dp[i][j]:=gcd(i,j);
    readln(n);
    sum:=0;
    for i:=1 to n do 
        for j:=1 to n do 
            for k:=1 to n do sum:=sum+dp[dp[i][j]][k];
    writeln(sum);
end.