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<iomanip>
using namespace std;
int main(){
  int n;
  cin>>n;
  int t[n],s[n],g[n],v[n];
  s[0]=g[0]=0;
  for(int i=1;i<=n; i++){
    cin >> t[i];
    s[i]=g[i-1];
    g[i]=s[i]+2*t[i];
  }
  for(int i=1;i<=n; i++){
    cin >> v[i];
  }
  int a=g[n];
  int vmin[a];
  for(int i=1;i<=a ;i++){
    vmin[i]=min(i,a-i);
  }
  for(int i=1;i<=n ;i++){
    for(int j=0;j<=s[i];j++){vmin[j]=min(vmin[j],s[i]-j+2*v[i]);}
    for(int j=s[i];j<=g[i];j++){vmin[j]=min(vmin[j],2*v[i]);}
    for(int j=g[i];j<=a;j++){vmin[j]=min(vmin[j],j-g[i]+2*v[i]);}
  }
  double ans =0;
  for(int i=1;i<=a ;i++){
    ans+=vmin[i];
  }
  cout <<fixed<<setprecision(3)<<ans/4<<endl;
  return 0;
}