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: Wrong Answer)

#include<stdio.h>

int main()
{
  int k, i, j, a, b, c, max_d;
  int d[] = {1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 
             41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};
  long sum = 0;
  
  scanf("%d", &k);
  
  for(a = 1; a <= k; a++)
  {
    for(b = 1; b <= k; b++)
    {
      for(c = 1; c <= k; c++)
      {
        for(j = 0; j < 26; j++)
        {
          if(a == b && b == c)    
            max_d = a;
          
          else if(a % d[j] == 0 && b % d[j] == 0 && c % d[j] == 0)
            max_d *= d[j];
        }
        sum += max_d;
        max_d = 1;
      }
    }
  }
  
  printf("%ld", sum);
  
  return 0;
}