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>
#include<math.h>
int gcd(long long int a,long long int b,long long int c)
{
   long long int n,m,t,j,k,h;
   if(a>b)
   {
       t=a;a=b;b=t;
   }
   if(a<=b)
   {
       m=b%a;
       if(m==0) n=a;
       while(m!=0)
       {
           b=a;
           a=m;
           m=b%a;
       }
       n=a;
   }

   if(n>c)
   {
       j=n;n=c;c=j;    //j×÷½»»»
   }
   if(n<=c)
   {
       k=c%n;
       if(k==0) h=n;
       while(k!=0)
       {
           c=n;
           n=k;
           k=c%n;
       }
       h=n;
   }
   return h;
}




int main()
{
    long long int n,i,j,k,sum=0;
    scanf("%lld",&n);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            for(k=1;k<=n;k++)
            {
                sum=sum+gcd(i,j,k);
            }
        }
    }
    printf("%lld",sum);
}