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>
 
using namespace std;
typedef long long ll;
typedef vector<int> VI;
ll mm=1000000000;ll MM=mm+7;
#define rep(i, n) for(int i=0;i<n;i++)
#define PI 3.141592653589793

int main(){
  int n;
  cin >> n;
  map<int,int> cnt;
  for(int x=1;x<=100;x++){
    for(int y=1;y<=100;y++){
      for(int z=1;z<=100;z++){
        int m=x*x+y*y+z*z+x*y+y*z+z*x;
        if(cnt.count(m)){
          cnt.at(m)++;
        }
        else{
          cnt[m]=1;
        }
      }
    }
  }
  rep(i,n){
    if(cnt.count(i+1))cout << cnt.at(i+1)<< endl;
    else cout << 0 << endl;
  }
}