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 <cmath>
#include <algorithm>
#include <string>
using namespace std;
#pragma warning (disable: 4996)

int n, t[109], v[109], s[109], p[40009];
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) { cin >> t[i]; t[i] *= 2; s[i + 1] = s[i] + t[i]; }
	for (int i = 1; i <= n; i++) { cin >> v[i]; v[i] *= 2; }
	for (int i = 0; i <= s[n + 1]; i++) {
		for (int j = 1; j <= n + 1; j++) { if (s[j] >= i) { p[i] = v[j - 1]; break; } }
		for (int j = 1; j <= n + 1; j++) {
			int dst = abs(s[j] - i);
			p[i] = min(p[i], dst + min(v[j - 1], v[j]));
		}
	}
	int sum = 0;
	for (int i = 0; i <= s[n + 1]; i++)sum += p[i];
	printf("%.12Lf\n", 0.25l*sum);
	return 0;
}