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
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <algorithm>
#include <string>
#include <utility>
#include <map>
#include <deque>
#define MAX_N 100
#define MAX_T 200
#define INF 1000000000000
typedef long long ll;
using namespace std;

int main(){
  ll n;
  ll l[102],r[102],v[102];
  ll rv[40000];
  double ans=0.0;
  cin>>n;
  
  l[0]=0;r[0]=0;
  for(int i=0;i<n;i++){
    int t;
    cin>>t;
    l[i+1]=r[i];
    r[i+1]=l[i+1]+2*t;
  }
  l[n+1]=r[n];r[n+1]=l[n+1];

  v[0]=0;v[n+1]=0;
  for(int i=0;i<n;i++){
    cin>>v[i+1];
    v[i+1]*=2;
  }
  for(int i=0;i<r[n+1];i++){
    rv[i]=200;
    for(int j=0;j<=n+1;j++){
      if(i<l[j]){
        rv[i]=min(rv[i],v[j]+l[j]-i);
      }else if(i>r[j]){
        rv[i]=min(rv[i], v[j]+i-r[j]);
      }else{
        rv[i]=min(rv[i],v[j]);
      }
    }
    ans+=rv[i];
  }
  printf("%.10lf\n",ans/4.0);
  return 0;
}
