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 <vector>
#include <stdio.h>
using namespace std;

int N;
double t[102],T[102] = {0};
//double dT[101] = {0};
double V[102];

double velocity(double l,double r,double x,double v){
	if(0<=x && x <= l) return v+l-x;
	else if(l<=x && x<=r) return v;
	else return v+x-r;
}

int main(){
	cin >> N;
	for(int i=1;i<=N;i++){
		cin >> t[i];
		T[i+1] = T[i] + t[i];
	}	
	for(int i=1;i<=N;i++){
		cin >> V[i];
	}
	V[0] = 0;
	V[N+1] = 0;
	double time = 0;
	vector<double> F;
	while(time<=T[N+1]){
		double nowv = 1e9;
		for(int i=0;i<=N+1;i++){
			if(i==0){
				nowv = min(nowv,velocity(0,0,time,0));
			}else if(i==N+1){
				nowv = min(nowv,velocity(T[N+1],T[N+1],time,0));
			}else{
				nowv = min(nowv,velocity(T[i],T[i+1],time,V[i]));
			}
		}
		F.push_back(nowv);
		time += 0.5;
	}
	double ans = 0;
	for(int i=0;i<F.size();i++){
		ans += 0.25*(F[i]+F[i+1]);
	}
	printf("%.6f\n",ans);
}