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<vector>
#include<iomanip>

using namespace std;

const int INF=100000008;

int main()
{
	int n;
	cin >> n;
	int T=0;
	vector<int> begi(n+2);
	vector<int> en(n+2);
	begi[0]=0;
	en[0]=0;
	for(int i=1;i<=n;i++)
	{
		int t;
		cin >> t;
		begi[i]=T;
		T+=t;
		en[i]=T;
	}
	begi[n+1]=T;
	en[n+1]=T;
	vector<int> v(n+2);
	v[0]=0;
	v[n+1]=0;
	for(int i=1;i<=n;i++)
		cin >> v[i];

	vector<double> velo(2*T+1,INF);

	for(int i=0;i<2*T+1;i++)
	{
		double now=i*0.5;
		for(int j=0;j<n+2;j++)
		{
			double lim;
			if(now<begi[j])lim=v[j]+begi[j]-now;
			else if(en[j]<now)lim=v[j]+now-en[j];
			else lim=v[j];
			if(velo[i]>lim)velo[i]=lim;
		}
	}

	double ans=0;

	for(int i=0;i<2*T;i++)
		ans+=(0.5*(velo[i]+velo[i+1])/2);

	cout << fixed << setprecision(13) << ans << endl;
}