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<bits/stdc++.h>
using namespace std;
int k,tsum;
double s,d[40100],a,b[40100],ans,n,t[100];
int main(void){
    cin>>n;
    for(int i=0;i<n;i++){cin>>t[i];tsum+=t[i];}
    for(int i=0;i<n;i++){
        cin>>a;
        for(int j=0;j<=t[i]*2;j++){
            if(j==0)d[k+j]=min(d[k+j],a);
            else d[k+j]=a;
        }
        k+=2*t[i];
    }
    d[2*tsum]=0;
    for(int i=0;i<2*tsum+1;i++){
        b[i+1]=min(b[i]+0.5,d[i+1]);
        //cout<<i<<d[i]<<endl;
    }
    b[2*tsum]=0;
    for(int i=2*tsum;i>0;i--)b[i-1]=min(b[i-1],min(b[i]+0.5,d[i-1]));
    for(int i=0;i<2*tsum;i++){
        //cout<<i<<" "<<b[i]<<endl;
        ans+=(b[i]+b[i+1]);
        //cout<<iac+1<<" "<<ans/4<<endl;
    }
    cout<<setprecision(10)<<ans/4<<endl;
}
