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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

N = int(input())


for i in range(1, N+1):
    sum = 0

    range_x = int((i-5)**0.5)+1 if i-5 >0 else 0
    s = set()
    for x in range(1, range_x):
        flag = False
        if x in s:
            continue
        range_y = int((i-x**2-x-3)**0.5)+1 if i-x**2-2*x-2 > 0 else 0
        for y in range(x, range_y):
            if y in s:
                continue
            range_z = int((i-x**2-y**2-x*y-x-y)**0.5)+1 if i-x**2-y**2-x*y-x-y > 0 else 0
            for z in range(y, range_z):
                if z in s:
                    continue
                if x**2 + y**2 + z**2 + x*y + y*z + z*x == i:
                    num = len(set([x,y,z]))
                    sum += 6 if num == 3 else 3 if num==2 else 1
                    s = s | set([x,y,z])
                    flag = True
                    break
            if flag:
                break

    print(sum)