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 (LuaJIT 2.1.0) (judge verdict: Accepted)

local K = io.read("*n")

local function euc(arg1, arg2)
    --print(arg1, arg2)
    if arg1 == arg2 or arg2 == 0 then
        return arg1
    elseif arg1 == 0 then
        return arg2
    elseif arg1 > arg2 then
        return euc(arg2, arg1 % arg2)
    else
        return euc(arg1, arg2 % arg1)
    end
end

local function gcd(i, j, k)
    return euc(euc(i, j), euc(j, k))
end

local total = 0
for i = 1, K do
    for j = 1, K do
        for k = 1, K do
            total = total + gcd(i, j, k)
        end
    end
end

print(total)