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: Runtime Error)

#include <stdio.h>
#include <math.h>

int main(){
  int i, j, k, N;
  int f, cnt[10050];
  
  scanf("%d", &N);
  
  for(i = 0; i <= 10050; i++) cnt[i];
  
  for(i = 1; i <= 105; i++){
    for(j = 1; j <= 105; j++){
      for(k = 1; k <= 105; k++){
        f = i*i + j*j + k*k + i*j + j*k + k*i;
        if(f<=10050){
          cnt[f]++;
        }
      }
    }
  }
  
  for(i = 0; i < N; i++) printf("%d\n", cnt[i+1]);
  
  return 0;
}