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;
import java.util.stream.IntStream;

/**
 * 解説どおりに実装したソースコード
 */
public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			int[] t = new int[n];
			IntStream.range(0, n).forEach(i -> t[i] = scanner.nextInt());
			// 時間制限区間の開始時間と終了時間
			int[] l = new int[n + 2], r = new int[n + 2];
			l[0] = r[0] = l[1] = 0;
			IntStream.range(0, n).forEach(i -> r[i + 1] = l[i + 2] = l[i + 1] + t[i]);
			r[n + 1] = l[n + 1];
			// 各区間の最大速度
			int[] v = new int[n + 2];
			v[0] = v[n + 1] = 0;
			IntStream.range(0, n).forEach(i -> v[i + 1] = scanner.nextInt());
			// 各時刻の最大速度(0.5秒刻み)
			double[] d = new double[r[n + 1] * 2 + 1];
			d[0] = 0.0d;
			System.out.println(IntStream.range(1, d.length).mapToDouble(i -> {
				d[i] = calcF(i, n, l, r, v);
				return (d[i - 1] + d[i]) / 4.0d;
			}).sum());
		}
	}

	/**
	 * 指定された時刻における走行可能の最大速度を取得する
	 * 
	 * @param index 時刻x2の値
	 * @param n     N
	 * @param l     時間帯のスタート時点
	 * @param r     時間帯のエンド時点
	 * @param v     時間帯の最大速度
	 * @return 指定された時刻における走行可能の最大速度
	 */
	private static double calcF(int index, int n, int[] l, int[] r, int[] v) {
		double time = index / 2.0d;
		return IntStream.range(0, n + 2).mapToDouble(i -> {
			if (time < l[i]) {
				return v[i] + (l[i] - time);
			} else if ((l[i] <= time) && (time <= r[i])) {
				return (double) v[i];
			} else {
				return v[i] + (time - r[i]);
			}
		}).min().getAsDouble();
	}
}