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 <string>
#include <map>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <complex>
#define rep(i, b) for(int i = 0; i < (b); i++)
#define repi(i, a, b) for(int i = (a); i < (b); i++)
#define repd(i, a, b) for(int i = (a); i >= (b); i--)
#define ALL(u) (u).begin(), (u).end()
#define INF 100000000

using namespace std;
typedef long long ll;

int main(){
	int n;
	while(cin >> n,n){
		int cnt = 0;
		for(int a = 0; a*a <= n; a++){
				for(int b = a; b*b <= n; b++){
						for(int c = b; c*c <= n; c++){
								for(int d = c; d*d <= n; d++){
									if(a*a + b*b + c*c + d*d == n) cnt++;
								}
						}
				}
		}
		cout << cnt << endl;
	}
	return 0;
}