Problem: A car moves from point A to point B at speed v meters per second. The action takes place on the X-axis. At the distance d meters from A there are traffic lights. Starting from time 0, for the first g seconds the green light is on, then for the following r seconds the red light is on, then again the green light is on for the g seconds, and so on.

The car can be instantly accelerated from 0 to v and vice versa, can instantly slow down from the v to 0. Consider that it passes the traffic lights at the green light instantly. If the car approaches the traffic lights at the moment when the red light has just turned on, it doesn't have time to pass it. But if it approaches the traffic lights at the moment when the green light has just turned on, it can move. The car leaves point A at the time 0.

What is the minimum time for the car to get from point A to point B without breaking the traffic rules?

Input

The first line contains integers l, d, v, g, r (1 ≤ l, d, v, g, r ≤ 1000, d < l) — the distance between A and B (in meters), the distance from A to the traffic lights, car's speed, the duration of green light and the duration of red light.

Output

Output a single number — the minimum time that the car needs to get from point A to point B. Your output must have relative or absolute error less than 10 - 6.

Examples

Input

2 1 3 4 5


Output

0.66666667


Input

5 4 3 1 1


Output

2.33333333

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class B implements Runnable {

    private static final boolean USE_FILE_IO = false;
    private static final String FILE_IN = "b.in";
    private static final String FILE_OUT = "b.out";

    BufferedReader in;
    PrintWriter out;
    StringTokenizer tokenizer = new StringTokenizer("");

    public static void main(String[] args) {
        new Thread(new B()).start();
    }


    private void solve() throws IOException {
        long l = nextLong();
        long d = nextLong();
        long v = nextLong();
        long g = nextLong();
        long r = nextLong();

        // the entire distance
        double answer = 1.0 * l / v;

        {
            // also add wait time
            long green = g * v;
            long red = r * v;
            long where = d % (green + red);

            if (where >= green) {
                where = green + red - where;
                answer += 1.0 * where / v;
            }
        }

        out.printf("%.12f", answer);
    }

    public void run() {
        long timeStart = System.currentTimeMillis();
        try {

            if (USE_FILE_IO) {
                in = new BufferedReader(new FileReader(FILE_IN));
                out = new PrintWriter(new FileWriter(FILE_OUT));
            } else {
                in = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(new OutputStreamWriter(System.out));
            }

            solve();

            in.close();
            out.close();
        } catch (IOException e) {
            throw new IllegalStateException(e);
        }
        long timeEnd = System.currentTimeMillis();
        System.out.println("Time spent: " + (timeEnd - timeStart) + " ms");
    }

    private String nextToken() throws IOException {
        while (!tokenizer.hasMoreTokens()) {
            String line = in.readLine();
            if (line == null) {
                return null;
            }
            tokenizer = new StringTokenizer(line);
        }
        return tokenizer.nextToken();
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private BigInteger nextBigInt() throws IOException {
        return new BigInteger(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

}