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:  Dooloper
*    created: 11.07.2020 21:06:39
**/

#include <bits/stdc++.h>

#define rep(i,n) for(int i = 0; i < (n); i++)
using namespace std;
using P = pair<int,int>;
using ll = long long;

const int mod = 1000000007;
class mint {
    long long x;
public:
    mint(long long x=0) : x((x%mod+mod)%mod) {}
    mint operator-() const {
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        return a;
    }
    // for prime mod
    mint inv() const {
        return pow(mod-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }

    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
};

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<ll> a(1010101);
  rep(i,n+1) a[i] = 0;
  for(int i = 1; i <= 100; i++){
    for(int j = 1; j <= 100; j++){
      for(int k = 1; k <= 100; k++){
        ll ans = 0;
        ans = i*i + j*j + k*k + i*j + j*k + k*i;
        a[ans]++;
      }
    }
  }
  for(int i = 1; i <= n; i++) cout << a[i] << endl;
  return 0;
}