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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class D5 implements Runnable {

	final double eps = 1e-9;

	private void Solution() throws IOException {
		double a = nextDouble(), v = nextDouble();
		double l = nextDouble(), d = nextDouble(), w = nextDouble();
		double t = 0;
		if (w + eps > v) {
			double s = v * v / (2 * a);
			if (s + eps > l)
				t = Math.sqrt(2 * l / a);
			else {
				double ta = v / a;
				double sa = a * ta * ta / 2;
				t = ta + (l - sa) / v;
			}
		} else {
			double sv = v * v / (2 * a);
			double sw = w * w / (2 * a);
			if (sw + eps > d) {
				if (sv + eps > l)
					t = Math.sqrt(2 * l / a);
				else {
					double ta = v / a;
					double sa = a * ta * ta / 2;
					t = ta + (l - sa) / v;
				}
			} else {
				double sd = (w * w - v * v) / (-2 * a);
				if (sv + sd < eps + d)
					t = v / a + (d - sv - sd) / v + (v - w) / a;
				else {
					double f = Math.sqrt(d * a + w * w / 2);
					t = f / a + (f - w) / a;
				}
				if (sd + eps > l - d) {
					double lv = Math.sqrt((l - d) * 2 * a + w * w);
					t += (lv - w) / a;
				} else {
					t += (v - w) / a + (l - d - sd) / v;
				}
			}
		}
		out.printf("%.12f", t);
	}

	public static void main(String[] args) {
		new D5().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer tokenizer;

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			Solution();
			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(0);
		}
	}

	void print(Object... obj) {
		for (int i = 0; i < obj.length; i++) {
			if (i != 0)
				out.print(" ");
			out.print(obj[i]);
		}
	}

	void println(Object... obj) {
		print(obj);
		print("\n");
	}

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

	long nextLong() throws NumberFormatException, IOException {
		return Long.parseLong(next());
	}

	double nextDouble() throws NumberFormatException, IOException {
		return Double.parseDouble(next());
	}

	String nextLine() throws IOException {
		return in.readLine();
	}

	String next() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens())
			tokenizer = new StringTokenizer(nextLine());
		return tokenizer.nextToken();
	}
}