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;

int main() {
  int n;
  cin>>n;
  int t[n];
  double max_v[n+1];
  for (int i=0; i<n; i++) cin>>t[i];
  for (int i=0; i<n; i++) cin>>max_v[i];
  max_v[n] = 0;

  double v[n+1];
  v[0] = 0;
  v[n] = 0;
  double cur_v = 0;
  for (int i=0; i<n-1; i++) {
    double max_vi = min(v[i] + t[i], max_v[i]);
    v[i+1] = max_vi;
    int acc_t = -t[i];
    for (int j=i+1; j<=n; j++) {
      acc_t += t[j-1];
      if (max_vi - acc_t <= max_v[j]) {
        continue;
      }
      v[i+1] = min(v[i+1], max_v[j] + acc_t);
    }
  }
  double ans = 0.0;
  for (int i=0; i<n; i++) {
    double v_start = v[i];
    double v_end = v[i+1];
    double v_high = (v_start + v_end + t[i]) / 2.0;
    v_high = min(v_high, max_v[i]);
    double t1 = v_high - v_start;
    double t2 = v_high - v_end;
    ans += (v_start + v_high) *t1 / 2.0 + (v_end + v_high) * t2 / 2.0 + v_high * (t[i] - t1 - t2);
  }
  printf("%.5lf", ans);
}