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<algorithm>
#include<iomanip>

using namespace std;

int main(){
	int n;
	cin >> n;
	vector<int> t(n + 1), v(n + 1);
	for(int i = 1; i <= n; i++){
		cin >> t[i];
	}
	for(int i = 1; i <= n; i++){
		cin >> v[i];
	}
	t.push_back(0);
	v.push_back(0);
	vector<int> l(1), r(1);
	for(int i = 1; i <= n + 1; i++){
		l.push_back(r[i - 1]);
		r.push_back(l[i] + t[i]);
	}
	vector<double> v2(1);
	for(double i = 0.5; i <= (double)r.back(); i += 0.5){
		double minv = 101.0;
		for(int j = 0; j < v.size(); j++){
			double tmp;
			if(i < l[j]){
				tmp = v[j] + l[j] - i;
			}else if(r[j] < i){
				tmp = v[j] + i - r[j];
			}else{
				tmp = v[j];
			}
			minv = min(minv, tmp);
		}
		v2.push_back(minv);
	}
	double ans = 0.0;
	for(int i = 0; i < v2.size() - 1; i++){
		ans += (v2[i] + v2[i + 1])  * 0.25;
	}
	cout << fixed << setprecision(5) << ans << endl;
	return 0;
}