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 <cstdio>
#include <vector>
#include <algorithm>
#include <numeric>

int main() {
  size_t N;
  scanf("%zu", &N);

  std::vector<int> t(N);
  int tsum = 0;
  for (auto& ti: t) {
    scanf("%d", &ti);
    ti += ti;
    tsum += ti;
  }

  std::vector<int> v(N);
  for (auto& vi: v) {
    scanf("%d", &vi);
    vi += vi;
  }

  std::vector<int> maxv{0};
  for (size_t i = 0; i < N; ++i) {
    if (maxv.back() > v[i]) maxv.back() = v[i];
    for (int j = 0; j < t[i]; ++j)
      maxv.push_back(v[i]);
  }
  maxv[tsum] = 0;

  std::vector<int> res(tsum+1);
  for (int i = 0; i <= tsum; ++i) {
    res[i] = 1e9;
    for (int j = 0; j < i; ++j) res[i] = std::min(res[i], res[j]+i-j);
    for (int j = i; j <= tsum; ++j) res[i] = std::min(res[i], maxv[j]+(j-i));
  }

  printf("%.12f\n", std::accumulate(res.begin(), res.end(), 0)/4.0);
}