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;

int dp[1<<20];
vector<int>a;
signed main(){
	for(int i=0;i<1<<8;i++){
		a.push_back((i+1)*(i+1));
		}
		for(int c=0;c<a.size();c++){
			dp[a[c]]++;
			}
			for(int c=0;c<a.size();c++){
			for(int d=c;d<a.size();d++){
			dp[a[c]+a[d]]++;
			}
			}
		for(int c=0;c<a.size();c++){
			for(int d=c;d<a.size();d++){
			for(int e=d;e<a.size();e++){
				dp[a[c]+a[d]+a[e]]++;
				}
			}
			}
			for(int c=0;c<a.size();c++){
			for(int d=c;d<a.size();d++){
			for(int e=d;e<a.size();e++){
				for(int f=e;f<a.size();f++){
					dp[a[c]+a[d]+a[e]+a[f]]++;
					}
				}
			}
			}
			int K;
			while(cin>>K,K)cout<<dp[K]<<endl;
		}