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>
#define ll long long
#define ld long double
#define pb push_back
#define sb __builtin_popcount
#define MOD (ll)1000000007
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
using namespace std;

int main()
{
     ios_base::sync_with_stdio(false);
     cin.tie(NULL);
     cout.tie(NULL);
     ll k;
     cin >> k;
     ll sum = 0;
     for (ll i = 1; i <= k; i++)
     {
          for (ll j = 1; j <= k; j++)
          {
               for (ll l = 1; l <= k; l++)
               {
                    sum += __gcd(__gcd(i, j), l);
               }
          }
     }
     cout << sum << endl;
}