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
import java.util.InputMismatchException;
import java.io.*;
import java.util.Locale;

/**
 * Generated by Contest helper plug-in
 * Actual solution is at the bottom
 */
public class Main {
    public static void main(String[] args) {
        InputReader in = new StreamInputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        run(in, out);
    }

    public static void run(InputReader in, PrintWriter out) {
        Solver solver = new Equation0();
        solver.solve(1, in, out);
        Exit.exit(in, out);
    }
}

class StreamInputReader extends InputReader {
    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar, numChars;

    public StreamInputReader(InputStream stream) {
        this.stream = stream;
    }

    public int read() {
        if (numChars == -1)
            throw new InputMismatchException();
        if (curChar >= numChars) {
            curChar = 0;
            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }

    @Override
    public void close() {
        try {
            stream.close();
        } catch (IOException ignored) {
        }
    }
}

abstract class InputReader {
    private boolean finished = false;

    public abstract int read();

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

    public String nextToken() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuffer res = new StringBuffer();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }

    private boolean isSpaceChar(int c) {
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    public void setFinished(boolean finished) {
        this.finished = finished;
    }

    public abstract void close();
}

interface Solver {
    public void solve(int testNumber, InputReader in, PrintWriter out);
}

class Exit {
    private Exit() {
    }

    public static void exit(InputReader in, PrintWriter out) {
        in.setFinished(true);
        in.close();
        out.close();
    }
}

class Equation0 implements Solver {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        // ax^2+bx+c
        // D=sqrt(b*b-4*a*c);
        long a = in.nextLong();
        long b = in.nextLong();
        long c = in.nextLong();
        if (a == 0) {
            solveLinear(b, c, out);
        } else {
            long d = b * b - 4 * a * c;
            if (d < 0) {
                out.println(0);
            } else if (d == 0) {
                out.println(1);
                out.printf(Locale.US, "%.10f\n", - (double) b / (2 * a));
            } else {
                out.println(2);
                double x0 = (-b - Math.sqrt((double) d)) / (2 * a);
                double x1 = (-b + Math.sqrt((double) d)) / (2 * a);
                out.printf(Locale.US, "%.10f\n", Math.min(x0, x1));
                out.printf(Locale.US, "%.10f\n", Math.max(x0, x1));
            }
        }
    }
    void solveLinear(long a, long b, PrintWriter out) {
        // a * x + b = 0
        if (a == 0 && b != 0) {
            out.println(0);
        } else if (a == 0) {
            out.println(-1);
        } else {
            out.println(1);
            out.printf(Locale.US, "%.10f\n", - (double) b / a);
        }
    }
}
