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>

using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
typedef pair<ll, ll> Pll;
typedef vector<int> Vi;
//typedef tuple<int, int, int> T;
#define FOR(i,s,x) for(int i=s;i<(int)(x);i++)
#define REP(i,x) FOR(i,0,x)
#define ALL(c) c.begin(), c.end()
#define DUMP( x ) cerr << #x << " = " << ( x ) << endl

const int dr[4] = {-1, 0, 1, 0};
const int dc[4] = {0, 1, 0, -1};

int counter[(1<<15)+1];

int main() {
  FOR(i, 1, 182) {
    counter[i*i]++;
    FOR(j, i, 182) {
      if (i*i+j*j > 1<<15) break;
      counter[i*i+j*j]++;
      FOR(k, j, 182) {
        if (i*i+j*j+k*k > 1<<15) break;
        counter[i*i+j*j+k*k]++;
        FOR(l, k, 182) {
          if (i*i+j*j+k*k+l*l > 1<<15) break;
          counter[i*i+j*j+k*k+l*l]++;
        }
      }
    }
  }

  while (1) {
    int n; scanf("%d", &n);
    if (not n) break;
    printf("%d\n", counter[n]);
  }
  return 0;
}