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>
#include <time.h>
int main(){
  int i,j,k,t=1;
  int N;
int ans=0;
  scanf("%d",&N);
  clock_t start,end;
  start = clock();
  for(i=1;i<=N;i++){
    t=i;
    for(j=1;j<=N;j++){
      for(k=1;k<=N;k++){
        if(k%t == 0 && j%t == 0 && i%t == 0){
          ans+=t;
        }else{
          t--;
        }
      }
    }
  }
  end = clock();
  printf("%f秒かかりました\n",(double)(end-start)/CLOCKS_PER_SEC);

  printf("%d\n",ans);
  return 0;
}