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)

// Created by Kshitij Anand NSIT
#include <bits/stdc++.h>
//#include <ext/numeric>
//using namespace __gnu_cxx;


using namespace std;


#define int long long
#define pb push_back
#define P pair<int,int>
#define F first
#define S second
#define vi vector<int>
#define vc vector<char>
#define vb vector<bool>
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()
#define mp(a, b) make_pair(a, b)
#define min3(a, b, c) min(min(a, b), c)
#define min4(a, b, c, d) min(min(a, b), min(c, d))
#define max3(a, b, c) max(max(a, b), c)
#define max4(a, b, c, d) max(max(a, b), max(c, d))
#define fill(arr,val) memset(arr,val,sizeof(arr))
#define inf 1e18
#define ps(x,y) fixed<<setprecision(y)<<x
#define db(x)  cout<<#x<<" : "<<x<<endl


const int N = 1000000007;

void solve(){

    int n,a,b,c, ans=0, count=0, sum=0;
    cin>>n;

    map<int , int> m;
    
    for(int i=1; i<=100; i++) {

        for(int j=1; j<=100; j++) {

            for(int k=1; k<=100; k++) {
                int val = i*i + j*j + k*k + i*j + j*k + i*k;
                m[val]++;
            }
        }
    }


    for(int i=1; i<=n; i++) {

        cout<<m[i]<<endl;
    }  
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    
    
  //  int t; cin>>t; while(t--)
    {     
      solve();
    }

  return 0;
}