Problem: The Department of economic development of IT City created a model of city development till year 2100.

To prepare report about growth perspectives it is required to get growth estimates from the model.

To get the growth estimates it is required to solve a quadratic equation. Since the Department of economic development of IT City creates realistic models only, that quadratic equation has a solution, moreover there are exactly two different real roots.

The greater of these roots corresponds to the optimistic scenario, the smaller one corresponds to the pessimistic one. Help to get these estimates, first the optimistic, then the pessimistic one.

Input

The only line of the input contains three integers a, b, c ( - 1000 ≤ a, b, c ≤ 1000) — the coefficients of ax2 + bx + c = 0 equation.

Output

In the first line output the greater of the equation roots, in the second line output the smaller one. Absolute or relative error should not be greater than 10 - 6.

Examples

Input

1 30 200


Output

-10.000000000000000
-20.000000000000000

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;

public class NewClass1 {
    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
        int[] A = iarray(in.readLine(), 3);
        int a = A[0];
        int b = A[1];
        int c = A[2];
        double D = (b*b)-(4*a*c);
        double r1 = (-b+Math.sqrt(D))/(2*a);
        double r2 = (-b-Math.sqrt(D))/(2*a);
        out.write(Math.max(r1, r2)+"\n");
        out.write(Math.min(r1, r2)+"\n");
        out.flush();
    }
    
    private static int iparse (String s) {
        int n = Integer.parseInt(s);
        return n;
    }
    
    private static long lparse (String s) {
        long n = Long.parseLong(s);
        return n;
    }
    
    private static double dparse (String s) {
        double n = Double.parseDouble(s);
        return n;
    }
    
    private static int[] iarray (String s, int n) {
        String[] s1 = s.split(" ");
        int[] A = new int[n];
        for (int i=0; i<n; i++) {
            A[i] = Integer.parseInt(s1[i]);
        }
        return A;
    }
    
    private static long[] larray (String s, int n) {
        String[] s1 = s.split(" ");
        long[] A = new long[n];
        for (int i=0; i<n; i++) {
            A[i] = Long.parseLong(s1[i]);
        }
        return A;
    }
    
    private static double[] darray (String s, int n) {
        String[] s1 = s.split(" ");
        double[] A = new double[n];
        for (int i=0; i<n; i++) {
            A[i] = Double.parseDouble(s1[i]);
        }
        return A;
    }
}