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>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>

#define REP(i,k,n) for(int i=k;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)

using namespace std;
typedef long long ll;

int main()
{
    int n;
    while(cin >> n && n) {
        int ans = 0;
        
        int m = sqrt(n) + 1;
        REP(i,1,m) {
            int a = i*i;
            if(a == n) ans++;
            REP(j,i,m) {
                int b = j*j;
                if(a + b == n) ans++;
                REP(k,j,m) {
                    int c = k*k;
                    if(a + b + c == n) ans++;
                    REP(l,k,m) {
                        int d = l*l;

                        if(a + b + c + d == n) ans++;
                    }
                }
            }
        }

        cout << ans << endl;
    }

    return 0;
}