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;
#define lp(i,n) for(int i=0;i<n;i++)
#define lpin(a,i,n) for(int i=0;i<n;i++){cin>>a[i];}
int coun[500000],coun2[500000],coun3[500000],coun4[500000];
int  main(){
  coun[1]=1;
  coun2[2]=1;
  coun3[3]=1;
  coun4[4]=1;
  for(int i=2;i*i<50000;i++){
    coun[i*i]++;
    for(int j=i*i;j<50000;j++){
      coun2[j]+=coun[j-i*i];
      coun3[j]+=coun2[j-i*i];
      coun4[j]+=coun3[j-i*i];
    }
  }
  while(1){
    int x;
    cin>>x;
    if(x==0) break;
    cout<<(coun[x]+coun2[x]+coun3[x]+coun4[x])<<endl;
  }
  return 0;
}