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<algorithm>
#include<iomanip>
using namespace std;

int main() {
	int n, T[100]; cin >> n;
	double V[100];
	for (int i = 0; i < n; i++)cin >> T[i];
	for (int i = 0; i < n; i++)cin >> V[i];
	
	int t = 0;
	double vl[40001], vr[40001];
	vl[0] = 0;
	for (int i = 0; i < n; i++) for (int j = 0; j < 2 * T[i]; j++) {
		vl[t + 1] = min(vl[t] + 0.5, V[i]);
		t++;
	}
	vr[t] = 0;
	for (int i = n - 1; i >= 0; i--) for (int j = 0; j < 2 * T[i]; j++) {
		vr[t - 1] = min(vr[t] + 0.5, V[i]);
		t--;
	}

	double ans = 0;
	t = 0;
	for (int i = 0; i < n; i++) for (int j = 0; j < 2 * T[i]; j++) {
		ans += (min(vl[t], vr[t]) + min(vl[t + 1], vr[t + 1])) / 4;
		t++;
	}
	cout << setprecision(9) << ans << endl;

	return 0;
}