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>
using namespace std;

constexpr int memo_row = 5;
constexpr int memo_lim = 1 << 15;
vector<vector<int>> memo(memo_row, vector<int>(memo_lim, 0));

int set_memo() {
    memo[0][0] = 1;

    for (int j = 1; j * j < memo_lim; ++j) {
        int jj = j * j;

        for (int k = 0; k < 4; ++k) {
            for (int l = 0; l + jj < memo_lim; ++l) {
                if (memo[k][l] != 0) {
                    memo[k + 1][l + jj] += memo[k][l];
                }
            }
        }
    }

    for (int j = 1; j < memo_lim; ++j) {
        memo[0][j] = memo[1][j] + memo[2][j] + memo[3][j] + memo[4][j];
    }

    return 0;
}

int main() {
    set_memo();

    int N;
    while (true) {
        cin >> N;
        if (N == 0) break;
        cout << memo[0][N] << endl;
    }
    return 0;
}