Problem: Everybody knows that the capital of Berland is connected to Bercouver (the Olympic capital) by a direct road. To improve the road's traffic capacity, there was placed just one traffic sign, limiting the maximum speed. Traffic signs in Berland are a bit peculiar, because they limit the speed only at that point on the road where they are placed. Right after passing the sign it is allowed to drive at any speed.

It is known that the car of an average Berland citizen has the acceleration (deceleration) speed of a km/h2, and has maximum speed of v km/h. The road has the length of l km, and the speed sign, limiting the speed to w km/h, is placed d km (1 ≤ d < l) away from the capital of Berland. The car has a zero speed at the beginning of the journey. Find the minimum time that an average Berland citizen will need to get from the capital to Bercouver, if he drives at the optimal speed.

The car can enter Bercouver at any speed.

Input

The first line of the input file contains two integer numbers a and v (1 ≤ a, v ≤ 10000). The second line contains three integer numbers l, d and w (2 ≤ l ≤ 10000; 1 ≤ d < l; 1 ≤ w ≤ 10000).

Output

Print the answer with at least five digits after the decimal point.

Examples

Input

1 1
2 1 3


Output

2.500000000000


Input

5 70
200 170 40


Output

8.965874696353

Solution (in JAVA):

Solution:
import java.util.*;

public class Rules {
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int a = in.nextInt();
    double maxSpeed = in.nextInt();
    double len = in.nextInt();
    double delayDist = in.nextInt();
    double delaySpeed = in.nextInt();
    
    // two cases: I reach the delayPoint at a speed higher than the delaySpeed, or I don't.
    double timeToDelayAtMax = travelS(a, 0.0, maxSpeed, delayDist);
    double timeToDelayAtDelay = travelS(a, 0.0, delaySpeed, delayDist);
    if (Math.abs(timeToDelayAtMax-timeToDelayAtDelay) < 0.00001) {
      // I can reach the delayPoint without being delayed
      double time = travelS(a, 0.0, maxSpeed, len);
      System.out.printf("%.9f\n", time);
      return;
    }
    
    // binary search to find best velocity to stop acceleration in beginning
    double lowV = delaySpeed;
    double highV = maxSpeed;
    int loopCount = 1000;
    double[] initial = null;
    double[] secondary = null;
    while (loopCount-->0) {
      double guessV = (lowV+highV)/2.0;
      initial = travelA(a, 0.0, guessV);
      secondary = travelA(a, guessV, Math.min(delaySpeed, maxSpeed));
      if (initial[1] + secondary[1] < delayDist) { // okay, we can go faster
        lowV = guessV;
      } else {
        highV = guessV;
      }
    }
    double totalTime = 0.0;
    double finalSpeed = 0.0;
    initial = travelA(a, 0.0, lowV);
    secondary = travelA(a, lowV, delaySpeed);
    totalTime = initial[0] + secondary[0];
    double totalDist = initial[1] + secondary[1];
    totalTime += (delayDist-totalDist)/maxSpeed;
    
    // now we have delayDist to go, and we are at delaySpeed
    totalTime += travelS(a, delaySpeed, maxSpeed, len-delayDist);
    System.out.printf("%.9f\n", totalTime);
  }
  
  // [0] = time in h, [1] = dist travelled, in km
  // input units are in km/h^2, km/h, km/h
  public static double[] travelA(int a, double startSpeed, double endSpeed) {
    if (startSpeed > endSpeed)
      a = -a;
    
    double time = (endSpeed - startSpeed) / a;
    double dist = 0.5*a*time*time + startSpeed*time;
    return new double[] {time, dist};
  }
  
  // returns time it takes to travel dist, with given inputs
  public static double travelS(int a, double startSpeed, double maxSpeed, double dist) {
    double timeToMax = (maxSpeed - startSpeed) / a;
    double targetTime = (-startSpeed + Math.sqrt(startSpeed*startSpeed + 2*a*dist)) / a;
    if (targetTime < timeToMax)
      return targetTime;
    
    double partialDist = 0.5*timeToMax*timeToMax*a + startSpeed*timeToMax;
    double remainingDist = dist - partialDist;
    targetTime = remainingDist / maxSpeed;
    return targetTime + timeToMax;
  }
}