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<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
using namespace std;

int main(){
  vector<int> t,v;
  double now_t = 0;
  double now_v = 0;
  double len = 0;
  int n;
  t.push_back(0);
  v.push_back(0);
  cin >> n;
  for(int i = 0; i < n ; i++){
    int a;
    cin >> a;
    t.push_back(a + t[i]);
  }
  for(int i = 0; i < n ; i++){
    int a;
    cin >> a;
    v.push_back(a);
  }
  v.push_back(0);

  int count = 0;
  for(double i = 0; i <= t[n]; i += 0.5){
    double mini = 100;
    if(i >= t[count]) count++;
    for(int j = 0; j <= n; j++){
      if(mini > (i - t[j] + v[j]) && i - t[j] >= 0){
        mini = i - t[j] + v[j];
      }
      if(mini > (t[j] - i + v[j+1]) && t[j] - i >= 0){
        mini = t[j] - i + v[j+1];
      }
    }
    if(mini > v[count]){
      mini = v[count];
    }
    len += 0.25*(now_v + mini);
    now_v = mini;
  }
  printf("%.5lf\n", len);
}