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 <iomanip>
using namespace std;

int main() {
	int n;
	cin >> n;
	vector<int> t(n), v(n), b(n);
	for (int i = 0; i < n; i++) {
		cin >> t[i];
	}
	for (int i = 0; i < n; i++) {
		cin >> v[i];
	}
	b[n-1] = 0;
	for (int i = n-1; i > 0; i--) {
		b[i-1] = min({b[i] + t[i], v[i], v[i-1]});
	}

	int a = 0;
	double ans = 0;
	for (int i = 0; i < n; i++) {
		if (a+t[i] < b[i]) b[i] = min(b[i], a+t[i]);
		double x = double(b[i]+t[i]-a) / 2;
		ans += a*x + x*x/2 + b[i]*(t[i]-x) + (t[i]-x)*(t[i]-x)/2;
		if (a + x > v[i]) {
			double tmp = double(a+b[i]+t[i]) / 2 - v[i];
			ans -= tmp * tmp;
		}
		a = b[i];
	}
	cout << fixed << setprecision(15);
	cout << ans << endl;
	return 0;
}