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 <cstdio>
#include <iostream>
#include <cmath>
#include <ctype.h>
#include <string> 
#include <sstream>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <queue>
#include <utility>
#include <vector>
#include <set>
#include <iomanip>
  
using namespace std;

#define pi 3.141592653589793

int main()
{
	int result[32769][5] = {};
	for(int i = 0; i * i < 32769; i++){
		result[i * i][1]++;
		for(int j = i * i + 1; j < min(i * i * 4 + 1, 32769); j++){
			for(int k = 2; k < 5; k++){
				result[j][k] += result[j - i * i][k - 1];
			}
		}
	}
	int n;
	while(1){
		cin >> n;
		if(n == 0) break;
		int ans = 0;
		for(int i = 1; i < 5; i++){
			ans += result[n][i];
		}
		cout << ans << endl;
	}
	return 0;
}