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 <string>
using namespace std;
double speed(double a);

double t_time = 0.0;
int N, t[100], v[100];
int main(void) {
	double distance = 0.0;
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> t[i];
		t_time += t[i];
	}
	for (int i = 0; i < N; i++) {
		cin >> v[i];
	}

	for (double i = 0.0; i < t_time; i += 0.5) {
		distance += 0.25 * (speed(i) + speed(i + 0.5));
	}
	printf("%lf", distance);

	return 0;
}
double speed(double x) {
	double f[102], t_t = 0, f_tmp;
	f[0] = x;
	f_tmp = f[0];
	f[1] = t_time - x;
	if (f_tmp > f[1])	f_tmp = f[1];
	for (int i = 0; i < N; i++) {
		t_t += t[i];
		if (t_t - t[i] >= x) {
			f[i + 2] = v[i] + (t_t - t[i] - x);
		}
		else if (t_t - t[i] < x && t_t >= x) {
			f[i + 2] = v[i];
		}
		else if (t_t < x) {
			f[i + 2] = v[i] + (x - t_t);
		}
		if (f_tmp > f[i + 2])	f_tmp = f[i + 2];
	}
	return f_tmp;
}