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 <cstdio>
using namespace std;

#define REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define DEB 0
#define all(x) x.begin(), x.end()

//max:2192

unsigned short int dp[5][32770];

int main(){
  unsigned short int i,j,k;
  int n;
  for(i=1; i*i<(1<<15); i++){
    dp[1][i*i] = 1;
    for(j=i*i+1; j<(1<<15); j++){
      for(k=1; k<4; k++){
	dp[k+1][j] += dp[k][j-i*i];
      }
    }
  }
#if DEB
  int mx = 0;
  rep(i,1<<15){
    rep(j,5){
      mx = max(mx, dp[j][i]);
    }
  }
  printf("max:%d\n",mx);
#endif
  while(scanf("%d",&n),n){
    printf("%d\n",dp[1][n]+dp[2][n]+dp[3][n]+dp[4][n]);
  }
  return 0;
}