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 JAVA):

Solution:
import java.util.Scanner;

public class Main {
	public static void main(String[] args){
		Main main = new Main();
		main.solveD();
	}

	private void solveD() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] t = new int[N];
		int sum_t = 0;
		for (int i = 0; i < N; i++) {
			t[i] = sc.nextInt() * 2;
			sum_t += t[i];
		}
		int[] v = new int[N];
		for (int i = 0; i < N; i++) {
			v[i] = sc.nextInt() * 2;
		}
		int tmp_v = 0;
		int tmp_t = 0;
		int[] max_v = new int[sum_t + 1];
		for (int i = 0; i < N; i++) {
			max_v[tmp_t] = v[i] < tmp_v ? v[i] : tmp_v;
			tmp_v = max_v[tmp_t];
			for (int t_ind = 0; t_ind < t[i]; t_ind++) {
				max_v[++tmp_t] = v[i] < tmp_v + 1 ? v[i] : tmp_v + 1;
				tmp_v = max_v[tmp_t];
			}
		}
		tmp_v = 0;
		for (int i = N - 1; i >= 0; i--) {
			max_v[tmp_t] = v[i] < tmp_v ? v[i] : tmp_v;
			tmp_v = max_v[tmp_t];
			for (int t_ind = 0; t_ind < t[i]; t_ind++) {
				int back_v = v[i] < tmp_v + 1 ? v[i] : tmp_v + 1;
				--tmp_t;
				max_v[tmp_t] = max_v[tmp_t] < back_v ? max_v[tmp_t] : back_v;
				tmp_v = max_v[tmp_t];
			}
		}
		double answer = 0;
		for (int i = 0; i < sum_t; i++) {
			answer += (max_v[i] + max_v[i + 1]);
		}
		System.out.println(answer * 0.125);
	}
}