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 ll long long
#define MOD 998244353
#define endl "\n"
#define print(A,n) for(ll i=0;i<n;++i)cout<<A[i]<<' ';cout<<endl;
#define take(A,n) for(ll i=0;i<n;++i)cin>>A[i];
//    for(auto& it : m)
void fastio(){ios_base::sync_with_stdio(false);cin.tie(NULL);}
//  priority_queue<ll int> pq;
//  priority_queue <ll int, vector<ll int>, greater<ll int> > pq;
//  vector<ll int> v;
//  vector< pair<ll int,ll int> > v;
//  set<ll int>::iterator it=s.begin();
//  unordered_map<  pair<ll int,ll int>, ll int> mp;


ll int power(ll int x, ll int y, ll int p)  
{  
   ll int res = 1;     // Initialize result  
  
    x = x % p; // Update x if it is more than or  
                // equal to p 
   
    if (x == 0) return 0; // In case x is divisible by p; 
  
    while (y > 0)  
    {  
        // If y is odd, multiply x with result  
        if (y & 1)  
            res = (res*x) % p;  
  
        // y must be even now  
        y = y>>1; // y = y/2  
        x = (x*x) % p;  
    }  
    return res;  
}  

ll int modInverse(ll int a,ll int m) 
{ 
    ll int g = __gcd(a, m); 
   
        // If a and m are relatively prime, then modulo inverse 
        // is a^(m-2) mode m 
        
       return      power(a, m-2, m); 
   
} 


int main()
{
  fastio();
ll int T;
T=1;

while(T--)
{
  ll int n;
  cin>>n;
  ll int i,j,k;
  map<ll int,ll int> mp;
  for(i=1;i<=100;i++)
  {
    for(j=1;j<=100;j++)
    {
      for(k=1;k<=100;k++)
      {
      
          mp[i*i+j*j+k*k+i*j+j*k+k*i]++;
      }
    }
  }
  for(i=1;i<=n;i++)
  {
    cout<<mp[i]<<endl;
  }

  }
    return 0;
}