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 Rust (1.42.0) (judge verdict: Time Limit Exceeded)

use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {n: usize};
    let mut buf: Vec<usize> = vec![0, 0, 0];
    let mut result: Vec<usize> = vec![0; n];

    dfs(0, n, &mut buf, &mut result);

    for i in result.iter() {
        println!("{}", i)
    }
}

fn dfs(index: usize, n: usize, buf: &mut Vec<usize>, result: &mut Vec<usize>) {
    let mut count = 0;
    if index == 3 {
        let mut sum = buf[0] * buf[0]
            + buf[1] * buf[1]
            + buf[2] * buf[2]
            + buf[0] * buf[1]
            + buf[1] * buf[2]
            + buf[2] * buf[0];

        if sum <= n {
            for i in 1..n + 1 {
                if sum == i {
                    result[i - 1] += 1;
                    // println!("{:?}", buf);
                    // println!("{}", sum);
                    break;
                }
            }
        }
    } else {
        for i in 1..n / 2 {
            buf[index] = i;
            dfs(index + 1, n, buf, result);
        }
    }
}