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.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		final Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] time = new int[n + 3];
		int[] maxv = new int[n + 2];
		time[0] = 0;
		time[1] = 0;
		for (int i = 0, k = 2; i < n; i++, k++) {
			time[k] = sc.nextInt();
			time[k] += time[k-1];
		}
		time[n + 2] = time[n + 1];
		int totalTime = time[n+2];
		maxv[0] = 0;
		maxv[n + 1] = 0;
		for (int i = 1; i <= n; i++) {
			maxv[i] = sc.nextInt();
		}

//		System.out.println(Arrays.toString(time));
//		System.out.println(Arrays.toString(maxv));
//		System.out.println(totalTime);

		double[] maxV_Every_0_5_seconds = new double[totalTime * 2+1];
		Arrays.fill(maxV_Every_0_5_seconds, Integer.MAX_VALUE);
		for (double i = 0; i < totalTime * 2+1; i++) {
			for(int j = 0; j < n+2; j++) {
				//j個目の条件はtime[j+1]-time[j]の区間の最大速度はmax[v]
				if(i <= 2*time[j]) {
					maxV_Every_0_5_seconds[(int) i] = Math.min(maxV_Every_0_5_seconds[(int) i], maxv[j]+time[j]-(i/2));
				}
				if(i >= 2*time[j+1]) {
					maxV_Every_0_5_seconds[(int) i] = Math.min(maxV_Every_0_5_seconds[(int) i], maxv[j]-time[j+1]+(i/2));
				}
				if(i >= 2*time[j] && i <= 2*time[j+1]) {
					maxV_Every_0_5_seconds[(int) i] = Math.min(maxV_Every_0_5_seconds[(int) i], maxv[j]);
				}
			}
		}
//		System.out.println(Arrays.toString(maxV_Every_0_5_seconds));
		double ans = 0;
		for(int i = 0; i < totalTime*2; i++) {
			ans += 0.25 * (maxV_Every_0_5_seconds[i]+maxV_Every_0_5_seconds[i+1]);
		}
		System.out.println(ans);
	}
}