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;
typedef long long ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;

int T[222];
double V[222];
double M[44444];

int main(){
  int N; cin>>N;
  rep(i,44444) M[i]=101;
  rep(i,N) cin>>T[i];
  rep(i,N) cin>>V[i];
  int sum=0;
  rep(i,N){
    T[i]*=2;
    for(int j=sum;j<=sum+T[i];j++){
      M[j]=min(M[j],V[i]);
    }
    sum+=T[i];
  }
  M[0]=M[sum]=0;
  rep(i,N){
    rep(j,sum) M[j+1]=min(M[j+1],M[j]+0.5);
    for(int j=sum-1;j>=0;j--) M[j]=min(M[j],M[j+1]+0.5);
  }
  double ans=0;
  rep(i,sum) ans+=(M[i]+M[i+1]);
  cout<<setprecision(15);
  cout<<ans/4<<endl;
  return 0;
}