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<cmath>
using namespace std;
const int MAX_N = (1 << 16);

bool square[MAX_N] = {false};

int main(){
    for(int i = 1; i * i < MAX_N; i++){
        square[i * i] = true;
    }
    int n;
    while(cin >>n){
        if(n == 0) break;

        int cnt = 0, tmp;

        for(int i = 0; ; i++){
            tmp = n - (i * i);
            if(tmp < i * i) break;

            for(int j = i; ; j++){
                tmp = n - (i * i + j * j);
                if(tmp < j * j) break;

                for(int k = j; ; k++){
                    int tmp = n - (i * i + j * j + k * k);
                    if(tmp < k * k) break;

                    if(square[tmp]) cnt++;
                }
            }
        }
        cout <<cnt <<endl;
    }
    return 0;
}