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;
const int64_t INF = (1LL << 62);

int main()
{
  int N, t = 0;
  double v;
  cin >> N;
  vector<double> T(N), maxv(40010, INF);
  maxv[0] = 0;
  for (int i = 0; i < N; i++)
    cin >> T.at(i);
  for (int i = 0; i < N; i++)
  {
    cin >> v;
    for (int j = 0; j <= T.at(i) * 2; j++)
    {
      maxv.at(t + j) = min(v, maxv.at(t + j));
    }
    t += T.at(i) * 2;
  }
  maxv[t] = 0;
  for (int i = 1; i <= t; i++)
    maxv[i] = min(maxv[i], maxv[i - 1] + 0.5);
  for (int i = t - 1; i >= 0; i--)
    maxv[i] = min(maxv[i], maxv[i + 1] + 0.5);
  double ans = 0;
  for (int i = 1; i <= t; i++)
    ans += ((maxv[i - 1] + maxv[i]) * 0.5) / 2.0;
  cout << fixed << setprecision(15) << ans << endl;
}