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;
  vector<int> t(N+1), v(N+1);
  for (int i=0; i<N; i++) cin >> t[i];
  for (int i=0; i<N; i++) cin >> v[i];
  int tsum = 0;
  for (int i=0; i<N; i++) tsum += t[i];
  v[N] = 0;
  vector<double> speed(tsum*2+1,1e6);
  int tm = 0;
  for (int i=0; i<=N; i++){
    tm += t[i]*2;
    int ve = v[i+1];
    for (int j=tm; j>=0; j--){
      speed[j] = min(speed[j], ve + (tm-j)*0.5);
    }
  }
  tm = 0;
  for (int i=0; i<N; i++){
    for (int j=tm; j<=tm+2*t[i]; j++){
      speed[j] = min((double)v[i], speed[j]);
    }
    tm += t[i]*2;
  }
  double ans = 0;
  double befsp = 0;
  for (int i=1; i<=tsum*2; i++){
    // cout << befsp << endl;
    ans += (befsp+min(speed[i],befsp+0.5))*0.5*0.5;
    befsp = min(speed[i],befsp+0.5);
  }
  cout << fixed << setprecision(10);
  cout << ans << endl;
}