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 <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <cctype>
#include <complex>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <cassert>
#include <iomanip>

using namespace std;

#define pb push_back
typedef long long ll;
typedef complex<int> P;

const int NMAX = 1<<16;
vector<int> a;

void ini(){
	for(int i=0; i*i < NMAX ;i++) a.push_back(i*i);
	return;
}

int dfs(int num, int i2, int k){
	if(k==0){
		if(num == 0) return 1;
		return 0;
	}
	int res = 0;
	for(int i=i2; i<a.size() && num>=a[i] ;i++){
		res += dfs(num-a[i], i, k-1);
	}
	return res;
}

bool solve(){
	int n;
	cin>> n;
	if(!n) return false;
	
	cout<< dfs(n,0,4)<< endl;
	return true;
}

int main(){
	cout.setf(ios::fixed);
	cout.precision(10);
	
	ini();
	while(solve());
	return 0;
}