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 <iostream>
#include <iomanip>
using namespace std;

double min(double a, double b){
    return a<b?a:b;
}

int main(){
    int N;
    int t[102],v[102];
    cin >> N;

    t[0]=0;
    for(int i=1;i<=N;i++){
        cin >> t[i];
        if(i>1) t[i]=t[i]+t[i-1];
    }
    v[0]=v[N+1]=0;
    for(int i=1;i<=N;i++) cin >> v[i];

    double vmax[50000],vmaxR[50000];
    vmax[0]=0.0;
    for(int i=1;i<=N;i++){
        int cur_t=t[i-1]*2+1;
        for(int j=cur_t;j<=t[i]*2;j++){
            vmax[j]=min(v[i],vmax[j-1]+0.5);
        }
    }

    vmaxR[t[N]]=0;
    for(int i=N;i>=1;i--){
        int nt=t[i-1]*2;
        for(int j=t[i]*2-1;j>=nt;j--){
            vmaxR[j]=min(v[i],vmaxR[j+1]+0.5);
        }
    }

    for(int i=0;i<=t[N]*2;i++){
        vmax[i]=min(vmax[i],vmaxR[i]);
    }

    double x=0;
    for(int i=1;i<=t[N]*2;i++){
        x+=(vmax[i]+vmax[i-1])/4.0;
    }

    cout << fixed << setprecision(9) << x << endl;

    return 0;
}