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:
#define _USE_MATH_DEFINES

#include <cstdio>
#include <cstdlib>
#include <string.h>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>

using namespace std;

typedef pair<long long int, long long int> P;
long long int INF = 1e18;
long long int MOD = 1000000007;

using namespace std;

int cnt[210000] = {};

int main(){
	
	int N = 185;
	
	for(int i = 0; i < N; i++){
		for(int j = i; j < N; j++){
			for(int k = j; k < N; k++){
				for(int l = k; l < N; l++){
					cnt[i * i + j * j + k * k + l * l]++;
				}
			}
		}
	}
	
	while(true){
		int num;
		cin >> num;
		if(num == 0){
			break;
		}
		cout << cnt[num] << endl;
	}
	
	return 0;
}