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 hi;
var 
    a, b, c, k, i, ans: longint;
    r: array[1..200] of longint;

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

begin
    readln(k);
    for i := 1 to k do 
        r[i] := i;
    ans := 0;
    for c := 1 to k do 
        for b := 1 to k do 
            for a := 1 to k do 
                ans := ans + gcd(gcd(r[a], r[b]), r[c]);
    writeln(ans);
end.