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;

float ts[102],te[102],vmax[102];
float v[40005];


float f(int j, float x){
  if(x<=ts[j]){
    return vmax[j]+(ts[j]-x);
  }else if(ts[j]<x&&x<=te[j]){
    return vmax[j];
  }else{
    return vmax[j]+(x-te[j]);
  }
}


int main(){
  int N;
  cin>>N;
  ts[0]=0;
  te[0]=0;
  for(int i=1;i<=N;i++){
    int t;
    cin>>t;
    ts[i]=te[i-1];
    te[i]=ts[i]+t;
  //  cout<<"ts["<<i<<"]="<<ts[i]<<endl;
  //  cout<<"te["<<i<<"]="<<te[i]<<endl;
  }
  ts[N+1]=te[N];
  te[N+1]=ts[N+1];
  int T=te[N+1];
  for(int i=1;i<=N;i++){
    cin>>vmax[i];
  }
  vmax[0]=0.0;
  vmax[N+1]=0.0;
  float INF=10000000000000.0;
  fill(v,v+2*T,INF);
  for(int i=0;i<=2*T;i++){
    for(int j=0;j<N+2;j++){
      v[i]=min(v[i],f(j,i/2.0));
    }
    //cout<<"v["<<i<<"]="<<v[i]<<endl;
  }
  float ans=0.0;
  for(int i=1;i<=2*T;i++){
    ans+=0.25*(v[i]+v[i-1]);
  }
  printf("%f\n",ans);
  return 0;
}