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 vector<int> vi;



int main(){
  vi dw;
  int dp[100000][5]={};
  for(int i=1;i*i<=100000;i++){
    dw.push_back(i*i);
  }

  for(int t=0;t<dw.size();t++){
    for(int i=1;i+dw[t]<100000;i++){
      for(int h=1;h<4;h++){
	if(i==dw[t] && h==1){dp[i][h]++;}
	dp[i+dw[t]][h+1]+=dp[i][h];
      }
    }
  }
  /*
  for(int i=1;i<26;i++){
    for(int t=1;t<=4;t++){
      cout<<dp[i][t]<<" ";
    }
    cout<<endl;
  }
  */
  while(1){
  int n;
  cin>>n;
  if(n==0){break;}
  cout<<dp[n][1]+dp[n][2]+dp[n][3]+dp[n][4]<<endl;
  }
  
  return 0;
}