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 Ruby (2.7.1) (judge verdict: Time Limit Exceeded)

n = gets.to_i
1.upto(n) do |i|
  cnt = 0
  sqrt = (Math.sqrt(i)).to_i+1
  (1..sqrt).each do |x|
    (1..sqrt).each do |y|
      #d = -3*x*x-2*x*y-3*y*y+4*i
      #next if d < 0
      #z = (0.5*(Math.sqrt(d)-x-y)).round
      ##(z.floor..z.ceil).each do |z|
      #  cnt += 1 if x*x+y*y+z*z+x*y+y*z+z*x==i #c+z*(b+z) == 0
      ##end
      (1..sqrt).each do |z|
        if x*x+y*y+z*z+x*y+y*z+z*x==i
          cnt += 1
          break
        end
      end
    end
  end
  p cnt
end