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 <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define ll long long
#define pb push_back
#define PII pair<int,int>
#define X first
#define Y second
#define MP make_pair
#define PI acos(-1.0)
#define eps 1e-8
const int maxn=5e4+10;
const int maxm=1.5e7+10;
const int mod=998244353;
using namespace std;
 



int main()
{
    int a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    int ans=0;
    for(int i=1;i<=a;i++)
        for(int j=1;j<=b;j++)
            for(int k=1;k<=c;k++)
                ans+=__gcd(i,__gcd(j,k));
    printf("%d",ans);
    // system("pause");
    return 0;
}