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<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
typedef pair<int, int> P;
typedef unsigned int ui;
#define stop char nyaa;cin>>nyaa;
int main() {
	int squ[32768] = {};
	for (int i = 1; i < 200; i++) {
		if (i*i >= 32768)break;
		else squ[i*i] = 1;
	}
	int n;
	while (cin >> n, n) {
		int cnt = 0;
		for (int i = 0; i < 200; i++) {
			for (int j = i; j < 200; j++) {
				for (int k = j; k < 200; k++) {
					int sum = i * i + j * j + k * k;
					if (sum > n)break;
					int res = n - sum;
					if (res >= k * k&&squ[res])cnt++;
				}
			}
		}
		cout << cnt << endl;
	}
	return 0;
}