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 <iomanip>
using namespace std;

int main() {
	int n;cin>>n;
	int t[n],v[n];
	int time=0;
	for(int i=0;i<n;i++){
		cin>>t[i];
		time+=t[i];
	}
	for(int i=0;i<n;i++)cin>>v[i];
	double speed[2*time+1];

	int k=1;
	for(int i=0;i<n;i++){
		if(i){
			if(v[i]<v[i-1])speed[k-1]=v[i];
		}
		for(int j=0;j<2*t[i];j++){
			speed[k]=v[i];
			k++;
		}
	}
	speed[0]=0;speed[2*time]=0;
	for(int i=1;i<2*time+1;i++){
		if(speed[i-1]+0.5<speed[i]){
			speed[i]=speed[i-1]+0.5;
		}
	}

	for(int i=2*time-1;i>0;i--){
		if(speed[i+1]+0.5<speed[i]){
			speed[i]=speed[i+1]+0.5;
		}
	}
	double ans=0;
	for(int i=1;i<2*time+1;i++){
		ans+=(speed[i-1]+speed[i])/4;
	}
	cout<<setprecision(15)<<ans<<endl;

		return 0;
}