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:
// D.

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(int argc, char *argv[])
{
	cout.precision(10);
	int n, elapsed = 0, speed = 0, x;
	double ans = 0;
	cin >> n;
	vector<int> t(n), v(n);
	for (int i = 0; i < n; ++i) {
		cin >> x;
		elapsed += x;
		t[i] = elapsed;
	}
	for (int i = 0; i < n; ++i) {
		cin >> v[i];
	}
	t.push_back(t.back());
	v.push_back(0);
	elapsed = 0;
	for (int i = 0; i <= n; ++i) {
		for (; elapsed < t[i]; ++elapsed) {
			int next = min(speed + 1, v[i]);
			for (int k = i + 1; k <= n; ++k) {
				next = min(next, v[k] + t[k - 1] - elapsed - 1);
			}
			if (speed == next && speed < v[i]) {
				ans += speed + 0.25;
			} else {
				ans += (double)speed + 0.5 * (next - speed) * 0.5 * 0.5;
			}
			speed = next;
		}
	}
	cout << ans << endl;
	return 0;
}