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<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;
#include<iomanip>
int lim[21000*2];
int aa[110],bb[110];
int main(){
	int n;cin>>n;
	for(int i=0;i<n;i++)cin>>aa[i];
	for(int i=0;i<n;i++)cin>>bb[i];
	int ruiseki=0;
	lol(i,21000*2)lim[i]=i;
	lol(u,n){
		int a=aa[u],b=bb[u];
		a*=2;b*=2;
		a+=ruiseki;
		for(int i=0;i<21000*2;i++){
			if(i<ruiseki)lim[i]=min(lim[i],ruiseki-i+b);
			if(i>a)lim[i]=min(lim[i],i-a+b);
			if(ruiseki<=i&&i<=a)lim[i]=min(lim[i],b);
		}
		ruiseki=a;
	}
	lol(i,21000*2)lim[i]=min(lim[i],ruiseki-i);
	int ans=0;
	for(int i=0;i<ruiseki;i++){
		ans+=(lim[i]+lim[i+1]);
		//cout<<lim[i]<<" ";
	}
	double tmp=ans+0.0;
	tmp/=8.0;
	cout<<fixed<<setprecision(10)<<tmp<<endl;
	return 0;
}