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 Ruby (2.7.1) (judge verdict: Accepted)

def gcd(a, b)
   if b == 0
      return a
   else
      return gcd(b, a%b);
    end 
end

x = Array.new(201) {Array.new(201)}
k = gets.to_i
sum = 0

1.upto(200) do |a|
    a.upto(200) do |b|
        x[a][b] = gcd(a,b)
        x[b][a] = gcd(a,b)
    end
end

1.upto(k) do |a|
    1.upto(k) do |b|
        1.upto(k) do |c|
            sum += gcd(x[a][b], x[b][c])
        end
    end
end

puts sum