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 <bits/stdc++.h>
#define REP(i, a, n) for(ll i=a; i<n; i++)
#define RREP(i, a, n) for(ll i=n-1; i>=a; i--)
#define SIZE_N (2<<15)
typedef long long ll;
const ll mod =1e9+7;
const ll inf =1e18;
using namespace std;

ll n;
int main() {
    while(cin>>n, n){
        ll cnt=0;

        for(int i=1 ; i*i <= n ; i++){
            if(i*i == n) cnt++;
            for(int j=i ; i*i + j*j <= n ; j++){
                if(i*i + j*j == n) cnt++;
                for(int k=j ; i*i + j*j + k*k <= n ; k++){
                    if(i*i + j*j + k*k == n) cnt++;
                    for(int l=k ; i*i + j*j + k*k + l*l <= n ; l++){
                        if(i*i + j*j + k*k + l*l == n) cnt++;
                    }
                }
            }
        }

        cout<<cnt<<endl;
    }
    return 0;
}