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)

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <cmath>
#include <map>
using namespace std;

typedef long long ll;
typedef vector<int> vi;

const double PI=acos(-1.0);

int N;
void input()
{
    cin>>N;
}

void solve()
{
    map<int,int> M;
    for(int x=1;x*x<=N;++x){
        for(int y=1;y*y<=N;++y){
            for(int z=1;z*z<=N;++z){
                M[x*x+y*y+z*z+x*y+y*z+z*x]++;
            }
        }
    }
    for(int i=1;i<=N;++i){
        cout<<M[i]<<endl;
    }
}

int main()
{
    cin.tie();
    ios::sync_with_stdio(false);
    input();
    solve();
    return 0;
}