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>
#include <cmath>

using std::cin;
using std::cout;
using std::endl;

int dfs(int n, int sum, int k, int sn) {
	int ret = 0;
	if ( k < 0 ) {
		return 0;
	}
	else {
		if (sn == sum) {
			return 1;
		}
	}
	if (sn > sum || n == 0) {
		return 0;
	}
	if ( sn + (k * pow(n,2)) < sum ) {
		return 0;
	}

	ret += dfs(n, sum, k-1, sn+pow(n,2));
	ret += dfs(n-1, sum, k, sn);
	
	return ret;
}

int main(void) {
	int n;
	cin >> n;

	int lim,ret;
	bool find = false;
	while ( n != 0 ) {
		find = false;
		for (int i = 1; !(find); i++) {
			if ( !(pow(i,2) < n) ) {
				lim = i;
				find = true;
			}
		}
		ret = dfs(lim, n, 4, 0);
		cout << ret << endl;
		cin >> n;
	}
	return 0;
}