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 <algorithm>
#include <cstring>
#include <vector>

using namespace std;

int dp[200][5][(1<<15)+1];

int main(){
    int n;
    vector<int> sq;
    
    for(int i=1;i*i<=(1 << 15);++i){
        sq.push_back(i*i);
    }
    int sqn = sq.size();



    dp[0][0][0] = 1;

    for(int i=1;i<=sqn;++i){
        for(int j=0;j<=(1 << 15);++j){
            for(int k=0;k<=4;++k){
                for(int l=0;l<=4;++l){
                    if(k+l <= 4 && j+sq[i-1]*l <= (1 << 15)){
                        dp[i][k+l][j+sq[i-1]*l] += dp[i-1][k][j];
                    }
                }
            }
        }
    }
    
    while(cin >> n && n){
        int ret = 0;

        for(int i=0;i<=4;++i){
            ret += dp[sqn][i][n];
        }
        cout << ret << endl;
    }
}