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>
#include <algorithm>

int n, t[105], v[105];
long double maxv[40005];
int main() {
	std::cin >> n;
	for (int i = 0; i < n; ++i) {
		std::cin >> t[i];
		if (i != 0)
			t[i] += t[i - 1];
	}

	for (int i = 0; i < n; ++i)
		std::cin >> v[i];

	{
		long long tmp = 1;
		for (long long i = 0; i < n; i++) {
			while (tmp < 2 * t[i]) {
				maxv[tmp] = v[i];
				tmp++;
			}
			maxv[tmp] = std::min(v[i], v[i + 1]);
			tmp++;
		}
	}

	for (long long i = 0; i < 2 * t[n - 1]; ++i)
		maxv[i + 1] = std::min(maxv[i] + 0.5, maxv[i + 1]);

	for (long long i = 2 * t[n - 1]; i > 0; --i)
		maxv[i - 1] = std::min({ maxv[i - 1], maxv[i] + 0.5 });

	long double ans = 0;
	for (long long i = 0; i < 2 * t[n - 1]; ++i)
		ans += (maxv[i] + maxv[i + 1]) / 4;

	std::setprecision(15);
	std::cout << std::fixed << ans << std::endl;
	return 0;
}