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<string>
#include<iomanip>
#include<cmath>
#include<vector>
#include<algorithm>

using namespace std;

#define int long long
#define rep(i,n) for(int i = 0; i < (n); i++)
#define INF ((long long)1e18)
#define MOD ((int)1e9+7)
#define endl "\n"

#define yn(f) ((f)?"Yes":"No")
#define YN(f) ((f)?"YES":"NO")

#define MAX (1<<16)

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout<<fixed<<setprecision(10);
	
	static int sq[MAX];
	
	for(int i = 1; i*i < MAX; i++){
		sq[i*i] = i;
	}
	while(true){
		int N;
		int ans = 0;
		
		cin>>N;
		
		if(!N) break;
		
		for(int i = 1; i*i <= N; i++){
			if(i*i == N){ans++;break;}
			for(int j = i; i*i + j*j <= N; j++){
				if(j*j + i*i == N){ans++;break;}
				for(int k = j; k*k + i*i + j*j <= N; k++){
					int l = N - (k*k + i*i + j*j);
					int l2 = sq[l];
					if(!l || l2 >= k) ans++;
				}
			}
		}
		
		cout<<ans<<endl;
	}
	
	
	return 0;
}