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<vector>

using namespace std;

int main(){
	vector<int> square;
	for(int i = 1;i <= 1<<8;i++){
		if(i*i > 1<<15) break;
		square.push_back(i * i);
	}
	while(1){
		int n;
		cin >> n;
		if(n == 0) break;
		int ans = 0;
		for(int i = 0;i < square.size();i++){
			if(n == square[i]) ans++;
			if(n < square[i]) break;
			for(int j = i;j < square.size();j++){
				if(n == (square[i]+square[j])) ans++;
				if(n < (square[i]+square[j])) break;
				for(int k = j;k < square.size();k++){
					if(n == (square[i]+square[j]+square[k])) ans++;
					if(n < (square[i]+square[j]+square[k])) break;
					int rest = n - (square[i] + square[j] + square[k]);
					bool flag = false;
					for(int l = 1;l<rest;l++){
						if(l*l > rest) break;
						if(l*l == rest && l*l >= square[k]){
							flag = true;
							break;
						}
					}
					if(flag) ans++;
				}
			}
		}
		
		cout << ans << endl;
	}
	return 0;
}