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],T = 0;
  for(int i = 0;i < n;i++){
    cin >> t[i];
    T += t[i];
  }
  double v[2*T+1];
  for(int i = 0;i < 2*T+1;i++){
    v[i] = 1000000;
  }
  int tsum = 0;
  for(int i = 0;i < n;i++){
    double vc;
    cin >> vc;
    for(int j = tsum;j <= tsum + 2*t[i];j++){
      v[j] = min(v[j],vc);
    }
    tsum += 2*t[i];
  }
  v[0] = 0;
  v[2*T] = 0;
  for(int i = 1;i < 2*T+1;i++){
    v[i] = min(v[i-1]+0.5,v[i]);
  }
  for(int i = 2*T-1;i > 0;i--){
    v[i] = min(v[i+1]+0.5,v[i]);
  }
  for(int i = 0;i <= 2*T;i++){
    //cout << i/2 << ' ' <<  v[i] << endl;
  }
  double result = 0;
  for(int i = 0;i < 2*T;i++){
    result += (v[i] + v[i+1])*0.25;
  }
  printf("%.16f\n", result);
}