Problem: You are given an equation: 

Ax2 + Bx + C = 0. 

Your task is to find the number of distinct roots of the equation and print all of them in ascending order.

Input

The first line contains three integer numbers A, B and C ( - 105 ≤ A, B, C ≤ 105). Any coefficient may be equal to 0.

Output

In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.

Examples

Input

1 -5 6


Output

2
2.0000000000
3.0000000000

Solution (in JAVA):

Solution:
import java.util.*; 
public class Equation {
public static void main(String args[])
{
	Scanner scan = new Scanner(System.in); 
	
	// Values for Quadratic
	double a; 
	double b;
	double c; 
	
	a = scan.nextDouble();
	b = scan.nextDouble();
	c = scan.nextDouble(); 

	getRoots(a,b,c); 
}	



private static void getRoots(double a, double b, double c)
{	
	int numOfRoots = 0;
	double root1 = 0;
	double root2 = 0; 
// a is equal to 0
if(a == 0)
{
	if(b == 0 && c == 0)
	{
		System.out.println("-1");
	}
	
	else if(b == 0 && c != 0)
	{
		System.out.println("-0");
	}
	
	else
	{
		root1 = (-1 * (double)(c) / (double)(b));
		System.out.printf("1\n%5f",root1);
	}
	
}

	
// a is not equal to 0
else
{
	
	double quadratic =  Math.pow(b, 2) - (4 * a * c); 
	if(quadratic >= 0)
	{
		root1 = ((-1)*b) - Math.sqrt(quadratic);
		root1 /= (2 * a); 
		root2 = ((-1)*b) + Math.sqrt(quadratic);
		root2 /= (2 * a);
		if(root1 == root2)
		{
			System.out.printf("1\n%5f", root1);
		}
		
		else
		{
			if(root1 < root2)
			System.out.printf("2\n%5f\n%5f",root1,root2 );
			else
				System.out.printf("2\n%5f\n%5f",root2,root1);
		}
	}
	else
	{
		System.out.println("0s");
	}
}
}
}