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<bits/stdc++.h>
#define rep(i,a,...) for(int i = (a)*(strlen(#__VA_ARGS__)!=0);i<(int)(strlen(#__VA_ARGS__)?__VA_ARGS__:(a));++i)
#define per(i,a,...) for(int i = (strlen(#__VA_ARGS__)?__VA_ARGS__:(a))-1;i>=(int)(strlen(#__VA_ARGS__)?(a):0);--i)
#define foreach(i, n) for(auto &i:(n))
#define pii pair<int, int>
#define pll pair<long long, long long>
#define all(x) (x).begin(), (x).end()
#define bit(x) (1ll << (x))
#define lambda(RES_TYPE, ...) (function<RES_TYPE(__VA_ARGS__)>)[&](__VA_ARGS__) -> RES_TYPE
#define method(FUNC_NAME, RES_TYPE, ...) function<RES_TYPE(__VA_ARGS__)> FUNC_NAME = lambda(RES_TYPE, __VA_ARGS__)
using ll = long long;
//const ll MOD = (ll)1e9+7;
const ll MOD = 998244353;
const int INF = (ll)1e9+7;
const ll INFLL = (ll)1e18;
using namespace std;
template<class t>
using vvector = vector<vector<t>>;
template<class t>
using vvvector = vector<vector<vector<t>>>;
template<class t>
using priority_queuer = priority_queue<t, vector<t>, greater<t>>;
template<class t, class u> bool chmax(t &a, u b){if(a<b){a=b;return true;}return false;}
template<class t, class u> bool chmin(t &a, u b){if(a>b){a=b;return true;}return false;}
#ifdef DEBUG
#define debug(x) cout<<"LINE "<<__LINE__<<": "<<#x<<" = "<<x<<endl;
#else
#define debug(x) (void)0
#endif

namespace templates{
  ll modpow(ll x, ll b){
    ll res = 1;
    while(b){
      if(b&1)res = res * x % MOD;
      x = x * x % MOD;
      b>>=1;
    }
    return res;
  }

  ll modinv(ll x){
    return modpow(x, MOD-2);
  }

  bool was_output = false;
  template<class t>
  void output(t a){
    if(was_output)cout << " ";
    cout << a;
    was_output = true;
  }
  void outendl(){
    was_output = false;
    cout << endl;
  }
  ll in(){
    ll res;
    scanf("%lld", &res);
    return res;
  }

  template<class t>
  istream& operator>>(istream&is, vector<t>&x){
    for(auto &i:x)is >> i;
    return is;
  }

  template<class t, class u>
  istream& operator>>(istream&is, pair<t, u>&x){
    is >> x.first >> x.second;
    return is;
  }

  template<class t>
  void in(t&x){
    cin >> x;
  }

  template<class t>
  t in(){
    t res; cin >> res; return res;
  }

  template<class t>
  void out(t x){
    cout << x;
  }

  template<class t>
  vector<t> sorted(vector<t> line,function<bool(t,t)> comp=[](t a,t b){return a<b;}){
    sort(line.begin(),line.end(),comp);
    return line;
  }

  template<class t>
  vector<t> reversed(vector<t> line){
    reverse(line.begin(),line.end());
    return line;
  }
}

using namespace templates;

int main(){
  int n = in();
  vector<int> ans(n+1,0);
  method(func,int,vector<int> line){
    int res = 0;
    int n = line.size();
    rep(i,n){
      rep(j,i,n){
        res += line[i] * line[j];
      }
    }
    return res;
  };
  for(int i=1;func({i})<=n;++i){
    for(int j=1;func({i,j})<=n;++j){
      for(int k=1;func({i,j,k})<=n;++k){
        ++ans[func({i,j,k})];
      }
    }
  }
  rep(i,1,n+1){
    cout << ans[i] << endl;
  }
  return 0;
}