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>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define int long long
using namespace std;
typedef pair<int, int> P;

int dp[35001][6];

signed main(){
    dp[0][0] = 1;
    rep(i, 1, 257){
        rep(j, 1, 35001){
            rep(k, 0, 4){
                if(j - i*i >= 0){
                    dp[j][k + 1] += dp[j - i*i][k];
                }
            }
        }
    }
    rep(i, 0, 35000){
        rep(j, 0, 5){
            dp[i][5] += dp[i][j];
        }
    }
    /*rep(i, 0, 100){
        cout << i << ' ' << dp[i] << endl;
    }*/
    int n;
    while(cin >> n, n){
        cout << dp[n][5] << endl;
    }
}