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>
using namespace std;

#define  int        long long int
 #define    sp(x)                 setprecision(x)
#define    bp(x)           __builtin_popcount(x)
 #define  ld         long double
 #define  f(i,j,n)            for(int i = j; i <= n; i++)
#define  r(i,n,j)       for(int i = n; i >= j; i--)
 #define  all(container)       container.begin() , container.end()
#define  sz(container)   (int)container.size()
 #define  ff             first
#define  ss             second
 #define  pii      pair <int , int>
#define  sp(x)       setprecision(x)
 #define  endl    "\n"
#define  mod  1000000007
 #define  pb    push_back
#define  mp    make_pair
 #define  T     int ttt; cin >> ttt; while(ttt--)
#define fast ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL);

int power(int x, int y, int p)  
{  
    int res = 1;    
    x = x % p; 
    if (x == 0) return 0;
    while (y > 0)  
    {     
        if (y & 1)  
            res = (res*x) % p;  
        y = y>>1; 
        x = (x*x) % p;  
    }  
    return res; 
}
int a[1000005];
int32_t main()
{
    fast
    f(i,1,100)
        f(j,1,100)
            f(k,1,100)
                a[i*i +j*j +k*k +i*j +j*k +k*i]++;
    int n;
    cin>>n;
    f(i,1,n)
        cout<<a[i]<<endl;
}