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>
int gcd(int a, int b) 
{ 
    // Everything divides 0  
    if (a == 0) 
       return b; 
    if (b == 0) 
       return a; 
   
    // base case 
    if (a == b) 
        return a; 
   
    // a is greater 
    if (a > b) 
        return gcd(a-b, b); 
    return gcd(a, b-a); 
} 
int main()
{
	int a;
    scanf("%d", &a);
    int b = a;
    int c = a;
    long long int sum = 0;
    for(int i=1;i<=a;i++)
    {
        for(int j=1;j<=b;j++)
        {
            for(int k=1;k<=c;k++)
            {
              int x = gcd(j,k);
                x = gcd(x, i);
                sum += x;
            }
        }
    }
    printf("%lld", sum);
	return 0;
}