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 <algorithm>
#include <deque>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using ll = long long;

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];
  }
  vector<double> bounds(N + 1, 0);
  for (int i = N - 1; i >= 1; i--) {
    bounds[i] = min({bounds[i + 1] + t[i], v[i], v[i - 1]});
  }
  double res = 0, cur = 0;
  for (int i = 0; i < N; i++) {
    double nxt = min(cur + t[i], bounds[i + 1]);
    double x = (t[i] + nxt - cur) / 2.0;
    double a = min(v[i] - cur, x);
    double b = min(v[i] - nxt, t[i] - x);
    res += (cur + (cur + a)) * a / 2.0;
    res += (nxt + (nxt + b)) * b / 2.0;
    res += v[i] * (t[i] - a - b);
    cur = nxt;
  }
  printf("%.10f\n", res);
  return 0;
}