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 all(x) (x).begin(),(x).end()
const int mod=1000000007,MAX=100005;

double func(double a,double b,double t,double v){
    if(a<=t&&t<=b) return v;
    if(t<a) return v+(a-t);
    return v+(t-b);
}

int main(){
    
    int N;cin>>N;
    vector<double> T(N+1,0),V(N);
    for(int i=0;i<N;i++){
        cin>>T[i+1];
        T[i+1]+=T[i];
    }
    for(int i=0;i<N;i++){
        cin>>V[i];
    }
    
    vector<double> S(T[N]*2+1);
    int j=0;
    for(double t=0.0;t<=T[N];t+=0.5){
        double v=1000.0;
        for(int i=0;i<N;i++){
            v=min(v,func(T[i],T[i+1],t,V[i]));
        }
        v=min(v,t);
        v=min(v,T[N]-t);
        S[j]=v;
        j++;
    }
    double ans=0.0;
    for(int i=1;i<2*T[N]+1;i++){
        ans+=0.25*(S[i-1]+S[i]);
    }
    
    cout<<setprecision(15)<<ans<<endl;
}