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) {
		solve();
	}

	private static void solve() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] t = new int[N];
		int[] v = new int[N];

		for (int i = 0; i < N; i++) {
			t[i] = sc.nextInt();
		}

		for (int i = 0; i < N; i++) {
			v[i] = sc.nextInt();
		}
		sc.close();

		int totalTime = 0;
		for (int i = 0; i < N; i++) {
			totalTime += t[i];
		}

		int[] vmax = new int[totalTime * 2 + 1];
		vmax[0] = 0;

		int elapsed = 0;

		for (int i = 0; i < N; i++) {
			for (int j = elapsed * 2 + 1; j < (elapsed * 2 + 1) + (t[i] * 2); j++) {
				vmax[j] = v[i];
			}
			elapsed += t[i];
		}
		vmax[totalTime * 2] = 0;

		double[] vact = new double[totalTime * 2 + 1];
		vact[0] = 0;

		for (int i = 1; i < vact.length - 1; i++) {
			vact[i] = Math.min(vact[i - 1] + 0.5, vmax[i]);
		}

		vact[vact.length - 1] = 0;

		for (int i = vact.length - 2; i > 0; i--) {
			if (vmax[i] > vmax[i + 1] && vmax[i + 1] != 0) {
				vact[i] = Math.min(vact[i + 1] + 0.5, Math.min(vact[i], vmax[i + 1]));
			} else {
				vact[i] = Math.min(vact[i + 1] + 0.5, Math.min(vact[i], vmax[i]));
			}
		}

		double area = 0;
		for (int i = 0; i < vact.length - 1; i++) {
			area += (vact[i] + vact[i + 1]) * 0.5 / 2;
		}
		System.out.println(area);
	}
}