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 <cstdio>
#include <set>
#include <algorithm>

int main() {
  std::set<int> sq;
  for (int i=1; i*i<=(1<<15); ++i) sq.insert(i*i);

  auto end = sq.end();

  while (true) {
    int n;
    scanf("%d", &n);
    if (n == 0) break;

    int res = sq.count(n);
    for (auto i=sq.begin(); i!=end; ++i)
      for (auto j=i; j!=end; ++j)
        for (auto k=j; k!=end; ++k) {
          int x = n-*i-*j-*k;
          if (*k <= x && sq.count(x)) {
            ++res;
          }
        }

    for (auto i=sq.begin(); i!=end; ++i)
      for (auto j=i; j!=end; ++j) {
        int x = n-*i-*j;
        if (*j <= x && sq.count(x)) {
          ++res;
        }
      }

    for (auto i=sq.begin(); i!=end; ++i) {
      int x = n-*i;
      if (*i <= x && sq.count(x)) {
        ++res;
      }
    }

    printf("%d\n", res);
  }
}