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:
#define _USE_MATH_DEFINES
#define INF 0x3f3f3f3f
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <limits>
#include <map>
#include <string>
#include <cstring>
#include <set>
#include <deque>
#include <bitset>
#include <list>
#include <cctype>
#include <utility>

using namespace std;

typedef long long ll;
typedef pair <int,int> P;
typedef pair <int,P > PP;

int tx[] = {0,1,0,-1};
int ty[] = {-1,0,1,0};

static const double EPS = 1e-8;

int dp[1000000][5];

int main(){
	int n;
	while(~scanf("%d",&n)){
		if(n==0) break;

		memset(dp,0,sizeof(dp));

		dp[0][0] = 1;
		for(int i=1;i*i<=n;i++){
			for(int k=1;k<=4;k++){
				for(int j=1;j<=n;j++){
					if(j-i*i<0) continue;
					dp[j][k] += dp[j-i*i][k-1];
				}
			}
		}

		printf("%d\n",dp[n][4] + dp[n][3] + dp[n][2] + dp[n][1]);
	}
}