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 SumofGCD;
var
    k, mid, fin, h, i, j  : Longint;

function findgcd(x : longint; y : longint):longint;
begin 
    if (y = 0) then findgcd := x
        else findgcd := findgcd(y, x mod y); 
end;

begin
    readln(k);
    fin := 0;
    for h := 1 to k do begin
        for i := 1 to k do begin
            for j := 1 to k do begin
                mid := findgcd(i, j);
                fin += findgcd(h, mid)
            end;
        end;
    end;
    writeln(fin);
end.