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 <bits/stdc++.h>
typedef long long ll;
using namespace std;

const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define ALL(v) (v.begin(),v.end())
#define COUT(x) cout<<(x)<<endl

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define ALL(v) (v.begin(),v.end())
#define COUT(x) cout<<(x)<<endl

int gcd(int x,int y){
  if(y==0)return x;
  if(x<=y) swap(x,y);
  if(x%y==0) return y;
  else return(gcd(y,x%y));//最大公約数
}


int main(){
    int k;
    cin >> k;
    int ans = 0;
    for(int i=1;i<k+1;i++){
        for(int j=1;j<k+1;j++){
            for(int p=1;p<k+1;p++){
                ans += gcd(i,gcd(j,p));
            }
        }
    }
    COUT(ans);
    return 0;
}