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 <bits/stdc++.h>
using namespace std;

int main()
{
  int N, total = 0;
  cin >> N;
  vector<double> T(N), V(N);
  for (int i = 0; i < N; i++)
  {
    cin >> T.at(i);
    total += T.at(i);
  }
  for (int i = 0; i < N; i++)
  {
    cin >> V.at(i);
  }
  vector<double> maxV(total * 2 + 1, 10000);
  maxV[0] = 0, maxV[total * 2] = 0;
  int l = 0;
  for (int i = 0; i < N; i++)
  {
    for (int t = l; t <= l + T.at(i) * 2; t++)
    {
      maxV[t] = min(maxV[t], V.at(i));
    }
    l += T.at(i) * 2;
  }
  for (int t = 1; t <= total * 2; t++)
  {
    maxV[t] = min(maxV[t], maxV[t - 1] + 0.5);
  }
  for (int t = total * 2 - 1; t >= 0; t--)
  {
    maxV[t] = min(maxV[t], maxV[t + 1] + 0.5);
  }
  double ans = 0.0;
  for (int t = 0; t <= total * 2; t++)
  {
    ans += maxV[t] * 0.5;
  }
  cout << fixed << setprecision(10) << ans << endl;
}