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;
int n;
double t[101],v[101];
int main()
{
	double V=0;double ans=0;
	cin>>n;
	for(int i=0;i<n;i++)cin>>t[i];
	for(int i=0;i<n;i++)cin>>v[i];
	for(int i=0;i<n;i++)
	{
		double sto=v[i+1],ma=0;
		for(int j=n;--j>i;)ma=min(ma+t[j],v[j]);
		v[i+1]=ma;
		if(v[i]<=v[i+1])
		{
			if(v[i]-V<=t[i])
			{
				ans+=(v[i]-V)*(v[i]+V)/2+(t[i]-v[i]+V)*v[i];
				V=v[i];
			}
			else
			{
				ans+=t[i]*(V+V+t[i])/2;
				V=V+t[i];
			}
		}
		else
		{
			if(v[i+1]-V>t[i])
			{
				ans+=t[i]*(V+V+t[i])/2;
				V=V+t[i];
			}
			else if(v[i]-V+v[i]-v[i+1]<=t[i])
			{
				ans+=(v[i]-V)*(v[i]+V)/2+(v[i]-v[i+1])*(v[i]+v[i+1])/2+(t[i]-v[i]+V-v[i]+v[i+1])*v[i];
				V=v[i+1];
			}
			else
			{
				double m=(t[i]+V+v[i+1])/2;
				ans+=(m-V)*(m+V)/2+(m-v[i+1])*(m+v[i+1])/2;
				V=v[i+1];
			}
		}
		v[i+1]=sto;
	}
	cout<<fixed<<setprecision(9)<<ans<<endl;
}