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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<int,int> pint;

vector<int> vt;
int t[101],v[101];
double c_area(double h1,double h2,double w){
    return 0.5*(h1+h2)*w;
}
int main(){
    int n,tt;
    cin>>n;
    cin>>t[0];
    FOR(i,1,n){
        cin>>tt;
        t[i]=tt+t[i-1];
    }
    rep(i,n) cin>>v[i];
    double cv=0.0,ans=0.0,pre=0.0;
    int ct=0;
    v[n]=0;
    for(double i=0.5;i<=t[n-1];i+=0.5){
        cv+=0.5;
        if(i>t[ct]) ++ct;
        rep(j,n){
            if(i<=t[j]) cv=min(cv,v[j+1]+(t[j]-i));
        }
        cv=min(cv,(double)v[ct]);
        ans+=c_area(pre,cv,0.5);
        pre=cv;
    }
    cout<<fixed<<setprecision(10)<<ans<<endl;
    return 0;
}