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>
using namespace std;

int t[200],v[200],l[200],r[200],pt[200],d[50000];
	
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin >> t[i];
		t[i] *= 2;
		pt[i]=pt[i-1]+t[i];
	}
	for(int i=1;i<=n;i++)
	{
		cin >> v[i];
		v[i] *= 2;
		l[i]=min(l[i-1], v[i]-pt[i]);
	}
	r[n]=0;
	for(int i=n-1;i>=0;i--)
		r[i]=min(r[i+1], v[i+1]-(pt[n]-pt[i]));
	for(int i=1;i<=n;i++)
		for(int j=pt[i-1];j<=pt[i];j++)
		{
			d[j]=min(l[i-1]+j, r[i]+(pt[n]-j));
			d[j]=min(d[j], v[i]);
		}
	double ans=0;
	for(int i=0;i < pt[n];i++)
		ans+=d[i]+d[i+1];
	printf("%.15f\n",ans/8.0);
}