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<algorithm>
#include<string.h>
#include<string>
#include<math.h>
#include<stdlib.h>
#include<time.h>
#include<map>
#include<queue>
#include<set>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
int n,maxv[105],l[105],r[105],T,tt;
double ans,las,now;
inline double v(double t){
	double ans=1e50,te;
	fo(i,0,n+1){
		te=maxv[i];
		if (t<l[i]) te=maxv[i]+l[i]-t;
		if (t>r[i]) te=maxv[i]+t-r[i];
		if (te<ans) ans=te;
	}
	return ans;
}
int main(){
	scanf("%d",&n);
	fo(i,1,n){
		scanf("%d",&tt);
		l[i]=r[i-1];
		r[i]=l[i]+tt;
	}
	r[n+1]=l[n+1]=r[n];
	fo(i,1,n) scanf("%d",&maxv[i]);
	las=0;
	for(double t=0.5;t<=r[n];t+=0.5){
		now=v(t);
		ans+=0.25*(las+now);
		las=now;
	}
	printf("%.12f\n",ans);
	return 0;
}