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;
using ll=long long;
int main(){
  ll N; cin>>N;
  ll T[N],V[N],a=0,A[N];
  for(int i=0; i<N; i++){ll t; cin>>t; t*=2; T[i]=t; a+=t, A[i]=a;}
  for(int i=0; i<N; i++){ll v; cin>>v; v*=2; V[i]=v;}
  ll s=0, t=0, Z[a+1]; Z[0]=0, Z[a]=0;
  for(int i=1,j=0; i<a; i++){
    Z[i]=V[j];
    if(i==A[j]){
      j++;
      Z[i]=min(Z[i],V[j]);
    }
  }
  for(int i=1; i<=a; i++){
    Z[i]=min(Z[i],Z[i-1]+1);}
  for(int i=a-1; i>=0; i--){
    Z[i]=min(Z[i],Z[i+1]+1);
  }
//  for(int i=139; i<=a; i++){cout<<"i:"<<i<<" Z:"<<Z[i]<<endl;}
  double ans=0;
  for(int i=0; i<a; i++){
    ans+=Z[i];
    if(Z[i]<Z[i+1]) ans+=0.5;
    else if(Z[i]>Z[i+1]) ans-=0.5;
  }
  cout<<fixed<<setprecision(10)<<ans/4<<endl;
}