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 PHP (7.4.4) (judge verdict: Time Limit Exceeded)

<?php
// C
fscanf(STDIN, '%d', $K);
$ans = 0;
for ($i = 1; $i <= $K; $i++) {
    for ($j = 1; $j <= $K; $j++) {
        for ($k = 1; $k <= $K; $k++) {
            $ans += gcd(gcd($i, $j), $k);
        }
    }
}
echo $ans;
 
// 最大公約数（2つ）
function gcd($m, $n)
{
    if (!$n) return $m;
    return gcd($n, $m % $n);
}
 
// 最大公約数（3つ以上）
function gcdAll($arr)
{
    $gcd = array_pop($arr);
    foreach ($arr as $a) {
        $gcd = gcd($gcd, $a);
    }
    return $gcd;
}