Problem: The fact that any positive integer has a representation as the sum of at most four positive squares (i.e. squares of positive integers) is known as Lagrange’s Four-Square Theorem. The first published proof of the theorem was given by Joseph-Louis Lagrange in 1770. Your mission however is not to explain the original proof nor to discover a new proof but to show that the theorem holds for some specific numbers by counting how many such possible representations there are.

For a given positive integer n, you should report the number of all representations of n as the sum of at most four positive squares. The order of addition does not matter, e.g. you should consider 42 + 32 and 32 + 42 are the same representation.

For example, let’s check the case of 25. This integer has just three representations 12 +22 +22 +42 , 32 + 42 , and 52 . Thus you should report 3 in this case. Be careful not to count 42 + 32 and 32 + 42 separately.



Input

The input is composed of at most 255 lines, each containing a single positive integer less than 215 , followed by a line containing a single zero. The last line is not a part of the input data.

Output

The output should be composed of lines, each containing a single integer. No other characters should appear in the output.

The output integer corresponding to the input integer n is the number of all representations of n as the sum of at most four positive squares.

Example

Input

1
25
2003
211
20007
0


Output

1
3
48
7
738

Solution (in CPP):

Solution:
#include <iostream>

using namespace std;
int Memo[33001];
int pow2[201];
void solve()
{
	for(int i = 0; i < 200; ++i)
	{
		pow2[i] = i * i;
	}
	for(int i = 1; i < 200; ++i)
	{
		int n1 = pow2[i];
		if(n1 > 33000)
		{
			break;
		}
		Memo[n1]++;
		for(int j = i; j < 200; ++j)
		{
			int n2 = n1 + pow2[j];
			if(n2 > 33000)
			{
				break;
			}
			Memo[n2]++;
			for(int k = j; k < 200; ++k)
			{
				int n3 = n2 + pow2[k];
				if(n3 > 33000)
				{
					break;
				}
				Memo[n3]++;
				for(int l = k; l < 200; ++l)
				{
					int n4 = n3 + pow2[l];
					if(n4 > 33000)
					{
						break;
					}
					Memo[n4]++;
				}
			}
		}
	}
	int num;
	while(cin >> num, num)
	{
		cout << Memo[num] << endl;
	}
}

int main()
{
	solve();
	return(0);
}