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.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.fill;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.sort;

public class Main {
	public static void main(String[] args) throws IOException {
		new Thread(null, new Runnable() {
			public void run() {
				try {
					try {
						if (new File("input.txt").exists())
							System.setIn(new FileInputStream("input.txt"));
					} catch (SecurityException e) {}
					new Main().run();
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
		}, "1", 1L << 24).start(); 
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer st = new StringTokenizer("");
	
	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		
		int a = nextInt();
		int b = nextInt();
		int c = nextInt();
		
		if (a == 0) {
			if (b == 0) {
				out.println(c == 0 ? -1 : 0);
			} else {
				solevLE(b, c);
			}
		} else {
			solveQE(a, b, c);
		}
		
		out.close();
	}
	
	void solevLE(double a, double b) {
		out.println(1);
		out.printf(Locale.US, "%.12f%n", -b / a);
	}

	void solveQE(long a, long b, long c) {
		long D = b * b - 4L * a * c;
		if (D < 0) {
			out.println(0);
		} else if (D == 0) {
			out.println(1);
			out.printf(Locale.US, "%.12f%n", -b / (2.0 * a));
		} else {
			double[] roots = { (-b - sqrt(D)) / (2.0 * a), (-b + sqrt(D)) / (2.0 * a) };
			sort(roots);
			out.println(2);
			out.printf(Locale.US, "%.12f%n", roots[0]);
			out.printf(Locale.US, "%.12f%n", roots[1]);
		}
	}

	/*************************************************************** 
	 * Utility
	 **************************************************************/
	long b2mb(long b) {
		return b >> 20;
	}
	
	void checkMemory() {
		System.err.println(b2mb(Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory()) + "/" + b2mb(Runtime.getRuntime().totalMemory()) + " MB");
	}
	
	/*************************************************************** 
	 * Input 
	 **************************************************************/
	String nextToken() throws IOException {
		while (!st.hasMoreTokens())
			st = new StringTokenizer(in.readLine());
		return st.nextToken();
	}
	
	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}
	
	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}
	
	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
	
	String nextLine() throws IOException {
		st = new StringTokenizer("");
		return in.readLine();
	}
	
	boolean EOF() throws IOException {
		while (!st.hasMoreTokens()) {
			String s = in.readLine();
			if (s == null)
				return true;
			st = new StringTokenizer(s);
		}
		return false;
	}
}