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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        for (int t = 1; t <= N; t++) {
            int ans = solve(t);
            System.out.println(ans);
        }
    }
    private int solve(int t) {
        int ans = 0;
        for (int x = 1; x * x <= t; x++) {
            int remain1 = t - x * x;
            if (remain1 < 0) {
                continue;
            }
            for (int y = 1; y * y <= remain1; y++) {
                int remain2 = remain1 - y * y - x * y;
                if (remain2 < 0) {
                    continue;
                }
                for (int z = 1; z * z <= remain2; z++) {
                    int remain = remain2 - z * z - x * z - y * z;
                    if (remain == 0) {
                        ans += 1;
                    }
                }
            }
        }
        return ans;
    }
}