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>
#include <random>
#include <ctime>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//#pragma GCC optimize("-O3")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("avx2")
#define ll long long
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define pll pair<ll,ll>
#define ld double
#define pld pair<ld,ld>
#define pii pair<int,int>
#define sqr(a) ((a)*(a))
#define all(v) v.begin(),v.end()

using namespace std;
//using namespace __gnu_pbds;

ll MOD = 998244353;
const ll INF = 1e18;
const ld E = 1e-18;
const ld PI=acos(-1);
mt19937_64 rnd(time(0));
ll p1 = 31, p2 = 37, p3 = 7, p4 = rnd() % 100 + 1, md1 = MOD, md2 = 998244353, md3 = 998244357, md4 = rnd() % 1000000000;

//using ord_set=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

ll n, m, T, a[1000001];

ll binpow(ll h, ll r, ll md = MOD)
{
    ll l = 1;
    while (r)
    {
        if (r & 1) l *= h;
        h *= h;
        //h %= md;
        r /= 2;
    }
    return l;
}

ll gcd(ll x, ll y)
{
    if (x < y) swap(x, y);
    while (x && y)
    {
        x %= y;
        swap(x, y);
    }
    return x + y;
}

int main()
{
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    //freopen("capitals.in","r",stdin);
    //freopen("capitals.out","w",stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= 100; i++)
        for (int j = 1; j <= 100; j++)
            for (int u = 1; u <= 100; u++)
                a[i*i+j*j+u*u+i*j+i*u+j*u]++;
    for (int i = 1; i <= n; i++)
        cout << a[i] << "\n";
    return 0;
}

/*
3
4 2
1 1
1 0
3 2
2 2

3 1
1 1
1 0
3 2

3 2
1 1
1 0
3 2
*/