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 B20 {
	public static void main(String[] args){
		Scanner br = new Scanner(System.in);
		long a = br.nextInt();
		long b = br.nextInt();
		long c = br.nextInt();
		if(a == 0 && b  == 0){
			if(c == 0)
				System.out.println(-1);
			else
				System.out.println(0);
		}
		else if(a == 0){
			System.out.printf("%d\n%.8f\n", 1, -c*1.0/b);
		}
		else if(b == 0){
			if(c*-1.0/a < 0){
				System.out.println(0);
			}
			else{
				if(c == 0){
					System.out.println(1);
					System.out.printf("%.8f", 0.0);
				}
				else{
					double res = Math.sqrt(-c/a);
					System.out.println(2);
					System.out.printf("%.8f\n%.8f\n", -res, res);
				}
			}
		}
		else{
			if(b*b < 4*a*c){
				System.out.println(0);
			}
			else if(b*b == 4*a*c){
				double res = -b/(2.0*a);
				System.out.println(1);
				System.out.printf("%.8f\n", res);
			}
			else{
				double res1 = (-b + Math.sqrt(b*b-4*a*c))/(2.0*a);
				double res2 = (-b - Math.sqrt(b*b-4*a*c))/(2.0*a);
				if(res2 < res1){
					double temp = res1;
					res1 = res2;
					res2 = temp;
				}
				System.out.println(2);
				System.out.printf("%.8f\n%.8f\n", res1, res2);
			}
		}
	}
}