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 <algorithm>
#include <vector>
using namespace std;
 
const int MAX_N = 1<<15;
 
vector<int> sq;
 
void make() {
  for(int i = 0; i*i <= MAX_N; ++i) {
    sq.push_back(i*i);
  }
}
 
int solve(int n) {
  int res = 0;
  for(int i = 0; i < sq.size() && sq[i] <= n; ++i) {
    for(int j = i; j < sq.size() && sq[i]+sq[j] <= n; ++j) {
      for(int k = j; k < sq.size() && sq[i]+sq[j]+sq[k] <= n; ++k) {
        int t = n - (sq[i] + sq[j] + sq[k]);
        vector<int>::iterator it = lower_bound(sq.begin(), sq.end(), t);
        if(it == sq.end() || *it != t) continue;
        if(*it >= sq[k]) ++res;
      }
    }
  }
  return res;
}
 
int main() {
  make();
  int n;
  while((cin >> n) && n) cout << solve(n) << endl;
  return 0;
}