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;

int t[104]={0};
double f[40005]={0};

double calc(int a, int b, int n){
    int d = (b-a) / n;
    double acc = 0;
    for(int i=a;i<b;i+=d){
        acc += (f[i] + f[i+d]) * d / 2.0;
    }
    return acc;
}

int main(){
    int n;
    cin >> n;
    int total = 0;
    for(int i=0;i<n;++i){
        cin >> t[i];
        total += t[i];
    }
    vector<int> v;
    v.push_back(0);
    for(int i=0;i<n;++i){
        int tmp;
        cin >> tmp;
        v[v.size()-1] = min(v.back(), tmp);
        for(int j=1;j<=2*t[i];++j)v.push_back(tmp);
    }
    v[v.size()-1] = 0;

    for(int i=1;i<=total*2;++i) f[i] = min((double)v[i], f[i-1] + 0.5);
    for(int i=total*2-1;i>=0;--i) f[i] = min(f[i+1]+0.5, f[i]);

    printf("%.6f\n",calc(0, total*2, total*2)/2.0);

    return 0;
}