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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define REP2(i, n) for (int i = 1; i <= (int)(n); i++)
#define FOR(i, m, n) for (int i = m; i<n; i++)
#define SORT(x) sort(x.begin(), x.end())
#define REVE(x) reverse(x.begin(), x.end())
#define ALL(x) (x).begin(), (x).end()
#define SUM(x) accumulate(x.begin(), x.end(),0);
#define vint(v,n) vector<int> v(n); REP(i,n) scanf("%lld", &v[i]);
#define vstr(v,n) vector<string> v(n); REP(i,n) scanf("%lld", &v[i]);
int k=0,n=0, m=0, l=0,r=0,a=0, b=0, c=0, d=0,e=0,ans=0,ans2=0,x=0,y=0;
string s,t,S;
signed main() {
  ios_base::sync_with_stdio(false);
    cin >> n;
    REP(i,n){
        REP2(x,60){
            REP2(y,60){
                REP2(z,60){
                    if((x*(x+y)+y*(y+z)+z*(z+x))==i){
                        ans++;
                    }
                }
            }
        }
        cout <<ans<<endl;
        ans= 0;
    }
}