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;

vector<int> a;
typedef long long ll;
// int dp[1 << 15][1 << 8][5];

// idxより後でk個以内をとってnを作る方法は何通り
int dfs(int n, int idx, int k)
{
    /*int & res = dp[n][idx][k];
    if (res != -1)return res;
*/
    int res = 0;
    if (n == 0) res = 1;
    else {
        res = 0;
        if (k != 0)
        {
            for (int i = idx; i < a.size(); i++)
            {
                if (n - a[i] >= 0)
                {
                    res += dfs(n - a[i], i, k - 1);
                }
            }
        }
    }
    return res;
}

int main(){
    for (int i = 1; i*i <= 1 << 15; i++){
        a.push_back(i*i);
    }
    // memset(dp, -1, sizeof(dp));
    int n;
    while (cin>> n && n)
    {
        cout << dfs(n, 0, 4) << endl;
    }
}