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 <cstdio>

using namespace std;

int main()
{
	int n, t[101], v[101];
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> t[i];
	for (int i = 0; i < n; i++)
		cin >> v[i];
	t[n] = 1, v[n] = 0;
	int x = 1, s = 0, d = 0;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < t[i] * 2; j++, x++) {
			int m = 10000;
			for (int k = 0, T = 0; k <= n; T += t[k++]) {
				int l = k == i ? v[k] * 2 : k < i ? v[k] * 2 + x - T * 2 - t[k] * 2 : v[k] * 2 - x + T * 2;
				if (l < m) m = l;
			}
			if (m == s)
				d += s * 2;
			else if (m < s)
				d += s * 2 - 1, s--;
			else
				d += s * 2 + 1, s++;
		}
	}
	//cout << double(d) / 8 << endl;
	printf("%f\n", double(d) / 8);
	return 0;
}