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>
/*???????????????????????§??????????????°?????????????????????
???????????????????????????????????°??§dp[0][sq]???????????£??\???????????????
sq??\????????°??¨?????????????????????????????°???????¨???????????????????
2????????\????????¢????????????????????????????????§?????????????¨????????????°??????
*/
using namespace std;

const int N = 32770;
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
	int dp[4][N+1] = {};
	for(int i = 1;i*i <= N;i++){
		for(int j = 0;j < 3;j++){
			int sq = i*i;
			if(j == 0)dp[0][sq] = 1;
			for(int k = 1;k <= N;k++){
				if(k + sq >= N)continue;
				dp[j + 1][k + sq] += dp[j][k];
			}
		}
	}
	int n;
	while(cin >> n,n){
		cout << dp[0][n] + dp[1][n] + dp[2][n] + dp[3][n] << endl;
	}
  return 0;
}