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.*;
import java.io.*;
import java.math.BigInteger;
import java.util.regex.Pattern;
import javafx.util.Pair;
import sun.util.locale.StringTokenIterator;

public class Foh {

    static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
    static StringBuilder out = new StringBuilder();
    static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
    static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));

    static int I(String s) {
        return Integer.parseInt(s);
    }

    static long L(String s) {
        return Long.parseLong(s);
    }

    static double D(String s) {
        return Double.parseDouble(s);
    }

    static String[] SA() throws IOException {
        return bf.readLine().split(" ");
    }

    public static void main(String[] args) throws IOException {

        StringTokenizer in = new StringTokenizer(bf.readLine());
        long a = I(in.nextToken());
        long b = I(in.nextToken());
        long c = I(in.nextToken());
        if (a == 0 && b == 0&&c==0) {
            System.out.println("-1");
        } else  if (a == 0 && b == 0&&c!=0) {
            System.out.println("0");
        } else if (a == 0 && b != 0) {
            System.out.println("1");
            System.out.printf("%.5f%n", ((-c * 1.0) / (b * 1.0)));
        } else if (b == 0 && a != 0) {

            if ((a < 0 && c >= 0) || (c <= 0 && a > 0)) {
                if (Math.sqrt((-c * 1.0) / (a * 1.0)) != 0) {
                    System.out.println("2");
                    System.out.printf("%.5f%n", -Math.sqrt((-c * 1.0) / (a * 1.0)));
                    System.out.printf("%.5f%n", Math.sqrt((-c * 1.0) / (a * 1.0)));
                } else {
                    System.out.println("1");
                    System.out.printf("%.5f%n", Math.sqrt((-c * 1.0) / (a * 1.0)));
                }
            } else {
                System.out.println("0");
            }
        } else {

            if ((b * b) - (4 * a * c) < 0) {
                System.out.println(0);
            } else {

                
                double max = Math.max(((((-b) * 1.0) + Math.sqrt((b * 1.0 * b * 1.0) - (4.0 * a * 1.0 * c * 1.0))) / (2.0 * a)), (((-b) - Math.sqrt((b * 1.0 * b * 1.0) - (4.0 * a * 1.0 * c * 1.0))) / (2.0 * a)));
                double min = Math.min(((((-b) * 1.0) + Math.sqrt((b * 1.0 * b * 1.0) - (4.0 * a * 1.0 * c * 1.0))) / (2.0 * a)), (((-b) - Math.sqrt((b * 1.0 * b * 1.0) - (4.0 * a * 1.0 * c * 1.0))) / (2.0 * a)));
                if (( max == min)) {
                    System.out.println("1\n");
                    System.out.printf("%.5f%n", (min));

                }
                else{
                    System.out.println("2\n");
                    System.out.printf("%.5f%n", (min));

                    System.out.printf("%.5f%n", (max));
                }
            }
        }
    }
    
}