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<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#define MaxN 110
#define MaxV 0x3f3f3f3f
double v[MaxN];
double t[MaxN];
double cal(double x)
{
	return x*x/2;
}
double max(double a,double b)
{
	return a>b ? a:b;
 } 
 double min(double a,double b)
 {
 	return a<b ?a:b;
  } 
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
		scanf("%lf",&t[i]);
	for(int i=0;i<n;i++)
		scanf("%lf",&v[i]);
	double now_v=0;
	double sum=0;
	for(int i=0;i<n;i++)
	{
		double limit=MaxV;
		double times=0;
		for(int k=i+1;k<=n;k++)
		{
			limit=min(limit,times+v[k]);
			times+=t[k];
		}
		double up,down,bet;
		up=min((limit+t[i]-now_v)/2,v[i]-now_v);
		up=min(up,t[i]);
		down=max(now_v+up-limit,0);
		bet=t[i]-up-down;
		sum+=((now_v+up)*(up+bet)-cal(up));
		sum+=((now_v+up)*down-cal(down));
		now_v+=up;
		now_v-=down;
	}
	printf("%.15lf\n",sum);
	return 0;
 }