Problem:
Score : 300 points

Problem Statement

Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions:

1 \leq x,y,z

x^2 + y^2 + z^2 + xy + yz + zx = n

Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

Constraints

All values in input are integers.

1 \leq N \leq 10^4

Input

Input is given from Standard Input in the following format:

N

Output

Print N lines. The i-th line should contain the value f(i).

Sample Input 1

20

Sample Output 1

0
0
0
0
0
1
0
0
0
0
3
0
0
0
0
0
3
3
0
0

For n=6, only (1,1,1) satisfies both of the conditions. Thus, f(6) = 1.

For n=11, three triples, (1,1,2), (1,2,1), and (2,1,1), satisfy both of the conditions. Thus, f(6) = 3.

For n=17, three triples, (1,2,2), (2,1,2), and (2,2,1), satisfy both of the conditions. Thus, f(17) = 3.

For n=18, three triples, (1,1,3), (1,3,1), and (3,1,1), satisfy both of the conditions. Thus, f(18) = 3.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

/*Author - sambhu show(jgec) */
#include<bits/stdc++.h>
using namespace std;
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);
#define itc ll t;cin>>t;while(t--)
#define PI 3.141592653589793238
//ang=(PI*ang)/180.0 in radian
#define Max(a,b,c) max(a,max(b,c));
#define Min(a,b,c) min(a,min(b,c));
#define all(v) v.begin(),v.end()
#define ll long long int
#define ld long double
#define pb push_back
#define eb emplace_back
#define gcd(a,b) __gcd(a,b)
#define fi first
#define se second
#define endl "\n"
int dx[4] = {0, -1, 0, 1};
int dy[4] = {1, 0, -1, 0};
int XX[8] = { -1, -1, -1, 0, 0, 1, 1, 1 };
int YY[8] = { -1, 0, 1, -1, 1, -1, 0, 1 };
int solve(int num)
{
	int c = 0;
	for (int i = 1; i <= sqrt(num); i++)
	{
		for (int j = 1; j <= sqrt(num); j++)
		{
			for (int k = 1; k <= sqrt(num); k++)
			{
				int v = i * i + j * j + k * k + i * j + j * k + k * i;
				if (v == num)
					c++;
				if (v > num)
					break;
			}
		}
	}
	return c;
}
int main()
{
	fio
	int n;
	cin >> n;
	vector<int>ans;

	for (int i = 1; i <= n; i++)
	{
		cout << solve(i) << endl;
	}
	return 0;
}