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 <string>
#include <vector>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <functional>
#include <set>
#include <sstream>
#include <map>
#include <queue>
#include <stack>

using namespace std;

vector<long long> sq;

//int memo[1<<15][200][4];

int dfs(long long sum,int pos,int num)
{

	if(sum==0) return 1;
	else if(num==4 || sum<0) return 0;

//	if(memo[sum][pos][num]!=0) return memo[sum][pos][num];

	int res=0;
	for(int i=pos;sq[i]<=sum;i++) res+=dfs(sum-sq[i],i,num+1);

//	return memo[sum][pos][num]=res;
	return res;
}

int main()
{

	for(int i=1;i*i<(1<<15);i++) sq.push_back(i*i);

	long long n;
	while(cin>>n,n){
		cout<<dfs(n,0,0)<<endl;
	}

	return 0;
}