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 "bits/stdc++.h"
using namespace std;

int main(){
	int n;
	cin>>n;
	vector<double> ans(40000,200);
	
	vector<int> t(n);
	vector<int> v(n);
	int maxt=0;
	for(int i=0;i<n;i++){
		cin>>t[i];
		maxt+=t[i];
	}
	for(int i=0;i<n;i++)cin>>v[i];
	
	double speed=0;
	int time=0;
	int ttime=0;
	for(int i=0;i<n;i++){
		ttime=0;
		while(ttime<t[i]*2){
			ttime+=1;
			if(speed<v[i]){
				speed+=0.5;
			}else if(speed>v[i]){
				speed=v[i];
			}
			ans[time+ttime]=speed;
		}
		time+=ttime;
	}
	
	speed=0;
	time=maxt*2;
	ttime=0;
	for(int i=n-1;i>=0;i--){
		ttime=0;
		while(ttime<t[i]*2){
			ttime+=1;
			if(speed<v[i]){
				speed+=0.5;
			}else if(speed>v[i]){
				speed=v[i];
			}
			ans[time-ttime]=min(ans[time-ttime],speed);
		}
		time-=ttime;
	}
	double sum=0;
	for(int i=1;i<maxt*2;i++)sum+=ans[i];
	cout<< fixed << setprecision(3) <<sum/2<<endl;
}