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 <iostream>
#include <cstdio>

using namespace std;

const int maxn=500005;
typedef long long ll;

int read()
{
	int f=0; char c=getchar();
	while(c>'9' || c<'0') c=getchar();
	while(c<='9' && c>='0') {f=f*10+c-'0'; c=getchar();}
	return f;
}

int n;
int sum;

int gcd(int x, int y)
{
	if(y==0) return x;
	else return gcd(y,x%y);
}

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			for(int k=1;k<=n;k++)
			{
				sum+=gcd(i,gcd(j,k));
			}
		}
	}
	cout<<sum<<endl;
	return 0;
}