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)

#![allow(non_snake_case)]
#![allow(unused_imports)]

use proconio::{input, fastout};
use proconio::marker::*;
use whiteread::parse_line;
use std::collections::*;
use num::*;
use num_traits::*;
use superslice::*;
use std::ops::*;

use itertools::Itertools;
use itertools_num::ItertoolsNum;

#[fastout]
fn main() {
    input!{
        n: usize,
    }

    for i in 1..=n {
        let mut ans = 0;
        let x_bound = (i as f64).sqrt() as usize + 1;
        // dbg!(&x_bound);
        for x in 1..=x_bound {
            if i <= x * x {
                continue;
            }

            let y_bound = ((i - x*x) as f64).sqrt() as usize + 1;

            for y in 1..=y_bound {
                if i <= x * x + y * y + x * y {
                    continue;
                }

                let z_bound = ((i - x*x - y*y - x*y) as f64).sqrt() as usize + 1;

                for z in 1..=z_bound {
                    if x*x + y*y + z*z + x*y + y*z + z*x == i {
                        // println!("n: {}, x: {} y: {} z: {}", i, x, y, z);
                        ans += 1;
                    }
                }
            }
        }
        println!("{}", ans)
    }
}