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<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;
int main(){
    double s=0,n,v1[200001],v2[200001],T[200],V[200],ans=0;
    cin>>n;
    rep(i,n){
        cin>>T[i],s+=T[i];
        if(i>0) T[i]+=T[i-1];
    }
    rep(i,n) cin>>V[i];
    double t;
    int x=0;
    rep(i,2*s+1){
        t=double(i)/2;
        if(t==0) v1[i]=0;
        else v1[i]=min(v1[i-1]+0.5,V[x]);
        if(x<n-1&&t==T[x]) x++;
    }
    x=n-1;
    for(int i=2*s;i>=0;i--){
        t=double(i)/2;
        if(t==s) v2[i]=0;
        else v2[i]=min(v2[i+1]+0.5,V[x]);
        if(x>0&&t==T[x-1]) x--;
    }
    rep(i,2*s) ans+=(min(v1[i],v2[i])+min(v1[i+1],v2[i+1]))/4;
    cout<<setprecision(10)<<ans;
}