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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double a, v, l, d, w;
bool case1_(double &t) {
  double tl = sqrt(l * 2 / a);
  if (a * tl > v) return false;
  double td = sqrt(d * 2 / a);
  if (a * td > w) return false;
  t = tl;
  return true;
}
bool case2(double &t) {
  double tmax = v / a;
  double dmax = tmax * tmax * a / 2;
  if (dmax >= d) {
    double td = sqrt(d * 2 / a);
    if (a * td > w) return false;
    t = tmax + (l - dmax) / v;
    return true;
  } else {
    if (w < v) return false;
    t = tmax + (l - dmax) / v;
    return true;
  }
}
void solve(double &t) {
  double need = l - d;
  double tmax = (v - w) / a;
  if ((v + w) * tmax / 2 > need) {
    double x1 = (-w + sqrt(w * w + 2 * a * need)) / a;
    double x2 = (-w - sqrt(w * w + 2 * a * need)) / a;
    if (x2 < 0)
      t += x1;
    else
      t += x2;
  } else {
    t += tmax;
    need -= (w + v) * tmax / 2;
    t += need / v;
  }
}
bool case3(double &t) {
  double t1 = sqrt((2 * d + w * w / a) / 2 / a);
  double t2 = t1 - w / a;
  if (t1 * a > v) return false;
  t = t1 + t2;
  solve(t);
  return true;
}
void case4(double &t) {
  double t0 = v / a;
  double t2 = (v - w) / a;
  double t1 = (d - (a * t0 * t0 / 2) - ((v + w) * t2 / 2)) / v;
  t = t0 + t1 + t2;
  solve(t);
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(NULL);
  cout << (fixed) << setprecision(9);
  while (cin >> a >> v >> l >> d >> w) {
    if (w > v) w = v;
    double t;
    if (case1_(t))
      cout << t << endl;
    else if (case2(t))
      cout << t << endl;
    else if (case3(t))
      cout << t << endl;
    else {
      case4(t);
      cout << t << endl;
    }
  }
  return 0;
}