Problem: In the year 2168, AtCoder Inc., which is much larger than now, is starting a limited express train service called AtCoder Express.

In the plan developed by the president Takahashi, the trains will run as follows:

* A train will run for (t_1 + t_2 + t_3 + ... + t_N) seconds.
* In the first t_1 seconds, a train must run at a speed of at most v_1 m/s (meters per second). Similarly, in the subsequent t_2 seconds, a train must run at a speed of at most v_2 m/s, and so on.



According to the specifications of the trains, the acceleration of a train must be always within ±1m/s^2. Additionally, a train must stop at the beginning and the end of the run.

Find the maximum possible distance that a train can cover in the run.

Constraints

* 1 \leq N \leq 100
* 1 \leq t_i \leq 200
* 1 \leq v_i \leq 100
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N
t_1 t_2 t_3 … t_N
v_1 v_2 v_3 … v_N


Output

Print the maximum possible that a train can cover in the run.
Output is considered correct if its absolute difference from the judge's output is at most 10^{-3}.

Examples

Input

1
100
30


Output

2100.000000000000000


Input

2
60 50
34 38


Output

2632.000000000000000


Input

3
12 14 2
6 2 7


Output

76.000000000000000


Input

1
9
10


Output

20.250000000000000000


Input

10
64 55 27 35 76 119 7 18 49 100
29 19 31 39 27 48 41 87 55 70


Output

20291.000000000000

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <iomanip>
using namespace std;

int main() {
  int N;
  cin >> N;
  vector<double> t(N), v(N);
  for (int i = 0; i < N; i++) cin >> t[i];
  for (int i = 0; i < N; i++) cin >> v[i];

  t.insert(t.begin(), 0);
  for (int i = 0; i < N; i++) t[i+1] += t[i];
  t.insert(t.begin(), 0);
  t.insert(t.end(), t.back());

  v.insert(v.begin(), 0);
  v.insert(v.end(), 0);

  double prevSpeed = 0, ans = 0;
  for (double time = 0.5; time <= t.back(); time += 0.5) {
    double speed = 100;
    for (int i = 0; i <= N+1; i++) {
      if (time < t[i])
        speed = min(speed, v[i] + (t[i] - time));
      else if (time > t[i+1])
        speed = min(speed, v[i] + (time - t[i+1]));
      else
        speed = min(speed, v[i]);
    }
    ans += 0.25 * (prevSpeed + speed);
    prevSpeed = speed;
  }
  cout << fixed << setprecision(3) << ans << endl;
}