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.util.*;
public class TrafficLights {
    
    // I am sure that math involved here is a bit convoluted. Ill revisit this problem later.
    static void solve(){
        Scanner sc = new Scanner(System.in);
        // Tot distance, Distance to lights, Velocity, Green light time, Red light time
        double l=sc.nextDouble(),d=sc.nextDouble(),v=sc.nextDouble(),g=sc.nextDouble(),r=sc.nextDouble();
        double dist = l, e = 0; // e - elapsed time
        double t = (l/v); // Total time
        if(t<=g) e = t;
        else {
            double t1 = (d/v); // Time to get to signal
            double t2 = ((l-d)/v); // Time to get from signal to destination
            // By (t1%(g+r)) == 0, my reasoning is that if its zero than the light is green 
            // otherwise its red and we would have to calculate the elapsed time on the red light time left
            // Ok.. So we have to calculate the elapsed red light time differently for t1 depending on its 
            // relation to (r+g)
            // ((r+g)-t1) and (r-(t1-(int)t1)) are both time left over for red light to end.. A bit of java ugliness involved.
            e = (t1+t2+(((t1%(g+r))==0) ? 0 : (t1<(g+r)) ? ((r+g)-t1) : (r-(t1-(int)t1)))); 
        }
        System.out.println(e);
        sc.close();
    }
    
    public static void main(String args[]) {
        solve();
    }
}
// 1000 999 1 1 1
// 1000 1 1 1 1000

// 2 1 3 4 5