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.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		// 整数の入力(スペース改行関係なし)
		int n = sc.nextInt();
      	for (int i = 1; i <= n; i++) {
          int root = (int)Math.sqrt(i);
          int keisannkekka = 0;
          for (int x = 1; x <= root; x++) {
              for (int y = 1; y <= root ; y++) {
                  for (int z = 1; z <= root ; z++) {
                  	if((x*x)+(y*y)+(z*z)+(x*y)+(y*z)+(z*x)== i){
						keisannkekka += 1;
                  	}
                  }
              }
          }
          System.out.println(keisannkekka);
        }
      
			
		
	}
}