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;
using ll = long long;
using pii = pair<int, int>;

int N;
int t[110], v[110];
int l[110], r[110];

double calc(double x) {
  double res = 1e3;
  for (int i = 0; i < N + 2; i++) {
    double tmp = v[i];
    if (x < l[i]) tmp += abs(l[i] - x);
    else if (x > r[i]) tmp += abs(x - r[i]);
    res = min(res, tmp);
  }
  return res;
}

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> t[i];
  }
  for (int i = 0; i < N; i++) {
    cin >> v[i];
  }
  int sum = 0;
  for (int i = 0; i < N; i++) {
    l[i] = sum;
    sum += t[i];
    r[i] = sum;
  }
  l[N] = r[N] = 0;
  l[N + 1] = r[N + 1] = sum;
  v[N] = v[N + 1] = 0;
  double ans = 0;
  for (int i = 0; i < 2 * sum; i++) {
    double tmp = i / 2.0;
    ans += (calc(tmp) + calc(tmp + 1 / 2.0)) / 4.0;
  }
  cout << ans << endl;

  return 0;
}