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 <map>
#include <vector>
using namespace std;
typedef long long int ll;

double func(double a,double b,double t,double v){
	if(a<=t&&t<=b)return v;
	if(t<a)return v+(a-t);
	return v+(t-b);
}

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int n; cin >> n;
	vector<double> t(n+1,0),v(n);
	for(int i=0;i<n;i++){
		cin >> t[i+1];
		t[i+1]+=t[i];
	}
	for(int i=0;i<n;i++){
		cin >> v[i];
	}
	vector<double> s(t[n]*2+1);
	int j=0;
	for(double tt=0;tt<=t[n];tt+=0.5){
		double vv=1e9;
		for(int i=0;i<n;i++){
			vv=min(vv,func(t[i],t[i+1],tt,v[i]));
		}
		vv=min(vv,tt);
		vv=min(vv,t[n]-tt);
		s[j]=vv;
		j++;
	}
	double ans=0;
	for(int i=1;i<=2*t[n];i++){
		ans+=0.25*(s[i-1]+s[i]);
	}
	printf("%.8f\n",ans);
}