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>

using namespace std;

int main(void){
	while( 1 ) {
		int N;
		cin >> N;
		if( N == 0 ) break;
		int ans = 0;

		for(int i = 1; i * i <= N; i++) {
			if( i * i == N ) ans++;
		}

		for(int i = 1; i * i <= N; i++) {
			for(int j = i; i * i + j * j <= N; j++) {
				if( i * i + j * j == N ) ans++;
			}
		}

		for(int i = 1; i * i <= N; i++) {
			for(int j = i; i * i + j * j <= N; j++) {
				for(int k = j; k * k + j * j + i * i <= N; k++) {
					if( i * i + j * j + k * k == N ) ans++;
				}
			}
		}

		for(int i = 1; i * i <= N; i++) {
			for(int j = i; i * i + j * j <= N; j++) {
				for(int k = j; k * k + j * j + i * i <= N; k++) {
					for(int l = k; l * l + k * k + j * j + i * i <= N; l++) {
						if( i * i + j * j + k * k + l * l == N ) ans++;
					}
				}
			}
		}

		cout << ans << endl;

	}
    return 0;
}