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++ (Clang 10.0.0) (judge verdict: Accepted)

// include
#include <array>
#include <algorithm>
#include <set>
#include <cstdint>
#include <cmath>
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <string>
#include <unordered_map>


// pow(10, 15) 10の15乗

//  多倍長整数
#include <boost/multiprecision/cpp_int.hpp>
using big_int = boost::multiprecision::cpp_int;


std::uint64_t f(std::uint64_t x, std::uint64_t y, std::uint64_t z) {
  return (x * x) + (y * y) + (z * z) + (x * y) + (y * z) + (z * x);
}

// ============================================
int main() {
    int N;
    std::cin >> N;
    std::vector<int> v(N);

    int limit = std::numeric_limits<int>::max();;
    for (int x = 1; x <= N; x++) {
      for (int y = 1; y <= N; y++) {
        for (int z = 1; z <= N; z++) {
          if (x+y+z > limit) {
            break;
          }
          std::uint64_t a = f(x, y, z);
//std::cout << "x:" << x << " y:" << y << "z:" << z << " n=" << a << std::endl;          
          if (a > N) {
//            std::cout << "b" << std::endl;
            limit = x + y + z + 1;
            break;
          }
          a = a-1;
          v[a] += 1;
        }
      }
    }
   for (int i = 0; i < N; ++i) {
     std::cout << v[i] << std::endl;
   }
  
    return 0;
}