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;
long double xtimeof(long double dx, long double a, long double v0) {
  if (a == 0) return dx / v0;
  long double za = .5 * a;
  long double zb = v0;
  long double zc = -dx;
  long double delta = pow(zb, 2) - 4 * za * zc;
  delta = sqrt(delta);
  long double up1 = -zb + delta;
  long double up2 = -zb - delta;
  long double down = 2 * za;
  up1 /= down;
  up2 /= down;
  cerr << dx << ' ' << a << ' ' << v0 << ": " << za << ' ' << zb << ' ' << zc
       << ": " << delta << ' ' << up1 << ' ' << up2 << endl;
  if (up1 < 0) return up2;
  if (up2 < 0) return up1;
  return min(up1, up2);
}
long double distof(long double v1, long double v2, long double a) {
  return (pow(v2, 2) - pow(v1, 2)) / (2 * a);
}
long double get(long double dx, long double a, long double v0,
                long double vmax) {
  long double disl = distof(v0, vmax, a);
  if (disl <= dx) {
    cerr << "moving " << disl << "km with rising acc and " << dx - disl
         << "km with const speed" << endl;
    cerr << xtimeof(disl, a, v0) << endl << xtimeof(dx - disl, 0, vmax) << endl;
    return xtimeof(disl, a, v0) + xtimeof(dx - disl, 0, vmax);
  }
  return xtimeof(dx, a, v0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  long double a, v, l, d, w;
  cin >> a >> v >> l >> d >> w;
  (w) = min((w), (v));
  long double disl = distof(0, v, a);
  long double disr = distof(v, w, -a);
  long double ltime, rbegin;
  if (disl + disr < d) {
    cerr << "can reach vmax" << endl;
    cerr << "moving first " << disl << "km with positive accelration" << endl;
    cerr << "moving next " << disr << "km with negative accelration" << endl;
    ltime = xtimeof(disl, a, 0) + xtimeof(disr, -a, v) +
            xtimeof(d - (disl + disr), 0, v);
    rbegin = w;
  } else {
    cerr << "can NOT reach vmax" << endl;
    long double m;
    {
      long double hi = v;
      long double lo = 0;
      for (int i = 0; i < 1000; i++) {
        long double mid = (lo + hi) / 2;
        long double disl = distof(0, mid, a);
        long double disr = mid < w ? 0 : distof(mid, w, -a);
        if (disl + disr <= d)
          lo = mid;
        else
          hi = mid;
      }
      m = lo;
    }
    cerr << "reaching " << m << " as top speed" << endl;
    disl = distof(0, m, a);
    disr = m < w ? 0 : distof(m, w, -a);
    ltime = xtimeof(disl, a, 0) + xtimeof(disr, -a, m);
    rbegin = min(m, w);
  }
  cerr << "passed the speed limit point with speed " << rbegin << "km/h in "
       << ltime << "h" << endl;
  long double ans = ltime + get(l - d, a, rbegin, v);
  cout << fixed << ans << '\n';
  return 0;
}