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 <bits/stdc++.h>
using namespace std;

const int INF = 1e+9;
typedef pair<int, int> P;
typedef pair<int, P> pp;

int N, M;
vector<int> z;

int ans[50000];

int main() {
	for(int i = 0; i < 200; ++i){
		if(i * i <= 32768)
			z.push_back(i * i);
	}
	int sz = z.size();
	for(int i = 0; i < sz; ++i){
		if(z[i] * 4 > 32768)
			break;
		for(int j = i; j < sz; ++j){
			if(z[i] + z[j] * 3 > 32768)
				break;
			for(int k = j; k < sz; ++k){
				if(z[i] + z[j] + z[k] + z[k] > 32768)
					break;
				for(int l = k; l < sz; ++l){
					if(z[i] + z[j] + z[k] + z[l] > 32768)
						break;
					ans[z[i] + z[j] + z[k] + z[l]]++;
				}
			}
		}
	}
	while(cin >> N && N){
		printf("%d\n", ans[N]);
	}
	return 0;
}