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 <bits/stdc++.h>
using namespace std;

int main() {
	int n, t[100], v[100];
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> t[i];
	}
	for (int i = 0; i < n; i++) {
		cin >> v[i];
	}
	int a[102], b[102], m[102];
	a[0] = b[n] = 0;
	for (int i = 0; i < n; i++) {
		if (t[i] < v[i] - a[i]) {
			a[i + 1] = a[i] + t[i];
		} else {
			a[i + 1] = v[i];
		}
	}
	for (int i = n - 1; i >= 0; i--) {
		if (t[i] < v[i] - b[i + 1]) {
			b[i] = b[i + 1] + t[i];
		} else {
			b[i] = v[i];
		}
	}
	for (int i = 0; i < n + 1; i++) {
		m[i] = min(a[i], b[i]);
	}
	double s = 0;
	for (int i = 0; i < n; i++) {
		int l = v[i] - m[i];
		int r = v[i] - m[i + 1];
		double ss = t[i] * v[i] - l * l / 2.0 - r * r / 2.0;
		if (l + r > t[i]) {
			int d = l + r - t[i];
			ss += d * d / 4.0;
		}
		s += ss;
	}
	cout << fixed << setprecision(4) << s << endl;
	return 0;
}