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>
#include<vector>
using namespace std;

int n;
vector<int> t;
vector<int> v;

double func(double l,double r,double v,double x){
  if(x<=l){
    return v+(l-x);
  }else if(l<x && x<r){
    return v;
  }else{
    return v+(x-r);
  }
}

double getv(double x){
  double ret = func(0,0,0,x);
  int tt = 0;
  for(int i=0;i<n;++i){
    ret = min(ret, func(tt,tt+t[i],v[i],x));
    tt += t[i];
  }
  ret = min(ret, func(tt,tt,0,x));
  return ret;
}

int main(){
  int T = 0;
  cin>>n;
  for(int i=0;i<n;++i){
    int tmp;
    cin>>tmp;
    T += tmp;
    t.push_back(tmp);
  }
  for(int i=0;i<n;++i){
    int tmp;
    cin>>tmp;
    v.push_back(tmp);
  }

  double ret = 0;
  for(int i=0;i<2*T;++i){
    double a1 = i*0.5;
    double a2 = (i+1)*0.5;
    double v1 = getv(a1);
    double v2 = getv(a2);
    ret += 0.5*0.5*(v1+v2);
  }
  cout<<fixed<<setprecision(15)<<ret<<endl;
}