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 <time.h>
using namespace std;
#include <math.h>

// to increase precision
// std::cout<<std::fixed;
// 	std::cout<<std::setprecision(6)<<z;
//clock_t tStart = clock();
//printf("Time taken: %.9fs\n", (double)(clock() - tStart)/CLOCKS_PER_SEC);
#define maxn 200007

#define ll long long int

ll mod=1e9+7;



int main(){
	int n;
	cin>>n;
	for(int i=1;i<=min(5,n);i++)
		cout<<0<<endl;
	if(n<6){

	}
	else{
		map<int,int> mp;
		int h=sqrt(2*n);
		for(int x=2;x<=h;x++){
			for(int y=2;y<=h;y++){
				for(int z=2;z<=h;z++){
					int sum=x+y+z;
					if(sum%2)
						continue;
					sum/=2;
					int ans= x*x + y*y + z*z ;
					if(ans<=2*n && ans%2==0){
						//cout<<x<<" "<<y<<" "<<z<<" "<<ans<<endl;
						if(sum-x<1||sum-y<1||sum-z<1)
							continue;
						ans/=2;
						mp[ans]++;
					}
				}
			}
		}
		for(int i=6;i<=n;i++)
			cout<<mp[i]<<endl;
	}
}