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:
//
//  main.cpp
//  D - AtCoder Express
//
//  Created by wenhan on 2017/10/28.
//  Copyright © 2017年 wenhan. All rights reserved.
//

#include <cstdio>
#include <algorithm>
using namespace std;
double t[105],v[105];
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lf",&t[i]);
    for(int i=1;i<=n;i++)
        scanf("%lf",&v[i]);
    double cur=0,ans=0;
    for(int i=1;i<=n;i++){
        double a=0,b=0,c=0;
        double lim=0x3f3f;
        double tmp=0;
        for(int j=i+1;j<=n+1;j++){
            lim=min(lim, v[j]+tmp);
            tmp+=t[j];
        }
        a=min(min(t[i], v[i]-cur), (-cur+t[i]+lim)*0.5);
        b=max(double(0),cur+a-lim);
        c=t[i]-a-b;
        ans+=0.5*(cur*2+a)*a;
        ans+=(cur+a)*c;
        ans+=0.5*((cur+a)*2-b)*b;
        cur+=a;
        cur-=b;
    }
    printf("%lf\n",ans);
    return 0;
}