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<algorithm>
#include<iomanip>
using namespace std;
int n,t[111],v[111],v2[111];
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)cin>>t[i],t[i]*=2;
	for(int i=0;i<n;i++)cin>>v[i],v[i]*=2;
	for(int j=n;j--;)v2[j]=min({v[j],v[j+1],v2[j+1]+t[j+1]});
	int now=0;
	long ans=0;
	for(int i=0;i<n;i++)
	{
		//cout<<now<<" "<<v[i]<<" "<<v2[i]<<endl;
		int left=v[i]-now;
		int right=v[i]-v2[i];
		//cout<<left<<" "<<right<<endl;
		if(left+right<=t[i])
		{
			ans+=v[i]*t[i]-right*right/2-left*left/2;
		}
		else if(t[i]<left&&now+t[i]<=v2[i])
		{
			ans+=t[i]*t[i]/2+now*t[i];
		}
		else
		{
			int x=(v2[i]+t[i]-now)/2;
			//now+x==v2[i]+(t[i]-x)<=>2*x==v2[i]+t[i]-now
			long get=(x*x+(t[i]-x)*(t[i]-x))/2;
			long Min=min(now,v2[i]);
			get+=(now-Min)*x+(v2[i]-Min)*(t[i]-x);
			ans+=get+Min*t[i];
		}
		//cout<<ans<<endl<<endl;
		now=min(now+t[i],v2[i]);
	}
	cout<<fixed<<setprecision(15)<<(double)ans/4<<endl;
}