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 <algorithm>
#include <cstring>

using namespace std;

int main(){
    int n;
    while(cin>>n&&n!=0){
        int cnt=0;
        int sum=0;
        for(int i=0;i*i<=n;i++){
            sum+=i*i;
            for(int j=i;j*j+sum<=n;j++){
                sum+=j*j;
                for(int k=j;k*k+sum<=n;k++){
                    sum+=k*k;
                    for(int l=k;l*l+sum<=n;l++){
                        sum+=l*l;
                        if(sum==n)
                            cnt++;
                        sum-=l*l;
                    }
                    sum-=k*k;
                }
                sum-=j*j;
            }
            sum-=i*i;
        }
        cout<<cnt<<endl;
    }
    return 0;
}