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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
typedef long long int lld;

int gcd(int a, int b) {
  int r;
  while (1) {
    if (b == 0) {
      return a;
    }
    r = a % b;
    a = b;
    b = r;
  }
}

int main(void) {
  lld s = 0;
  int k, t;
  scanf("%d", &k);
  for (int a = 1; a <= k; a++) {
    if (a == 1) {
      s += k * k;
      continue;
    }
    for (int b = 1; b <= k; b++) {
      if (b == 1) {
        s += k;
        continue;
      }
      t = gcd(a, b);
      if (t == 1) {
        s += k;
        continue;
      }
      for (int c = 1; c <= k; c++) {
        if (c == 1) {
          s++;
          continue;
        }
        s += gcd(t, c);
      }
    }
  }
  printf("%lld\n", s);
}