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 Lua (Lua 5.3.5) (judge verdict: Time Limit Exceeded)

K=io.read()

function gcd(a,b)
    if b ~= 0 then
        return gcd(b, a % b)
    else
        return a
    end
end

sum = 0
for i=1,K do
    for j=1,K do
        for k=1,K do
            sum = sum + gcd(gcd(i,j),k)
        end
    end
end
print(sum)