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 JAVA):

Solution:
import java.util.ArrayList;
import java.util.Scanner;

public class Main {

	Scanner sc = new Scanner(System.in);
	public void run() {
		keisan();
		while(sc.hasNext()){
			int n = sc.nextInt();
			if(n == 0)break;
			else calc(n);
			
		}
	}
	int max = (int)Math.pow(2, 15);
	int m = (int)Math.sqrt(max) +1;
	int[] memo = new int[max + 1]; 
	public void keisan(){
		for(int i = 0; i < m; i++){
			for(int j = i; j < m; j++){
				for(int k = j; k < m; k++){
					for(int h = k; h < m; h++){
						if(i*i + j*j + k*k + h*h > max) break;
						memo[i*i + j*j + k*k + h*h] = memo[i*i + j*j + k*k + h*h] + 1;
					}
				}
			}
		}
	}
	public void calc(int n){
		System.out.println(memo[n]);
	}
	
	

	
	public static void main(String[] args) {
		new Main().run();
	}
}