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
#include <iostream>
#include <vector>
#include <iomanip>
#define REP(i, n) for (int i = 0; i < (n); i++)

using namespace std;

int main()
{
  int N; cin >> N;

  vector<int> t(N), v(N);
  int T = 0;
  REP(i, N) { cin >> t[i]; T += t[i]; }
  REP(i, N) cin >> v[i];

  vector<double> maxv(2*T + 1, 1e10);
  maxv[0] = maxv[2*T] = 0;
  int now = 0;
  REP(i, N) {
    for (int j = now; j <= now + 2*t[i]; j++) {
      maxv[j] = min(maxv[j], (double)v[i]);
    }
    now += 2*t[i];
  }

  for (int i = 1; i <= 2*T; i++) {
    maxv[i] = min(maxv[i], maxv[i-1] + 0.5);
  }
  for (int i = 2*T-1; i >= 0; i--) {
    maxv[i] = min(maxv[i], maxv[i+1] + 0.5);
  }

  //for (int i = 0; i <= 2*T; i++) {
  //  cout << (i / 2.0) << ',' << maxv[i] << endl;
  //}

  double ans = 0;
  REP(i, 2*T) {
    ans += (maxv[i] + maxv[i + 1])*0.25;
  }
  cout << fixed << setprecision(20) << ans << endl;

  return 0;
}