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.*;

public class Main {

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

	public void run() {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int t[] = new int[n];
		int v[] = new int[n];

		int sumt[] = new int[n+1];
		sumt[0]=0;
		for(int i=0; i<n; i++) {
			t[i]=sc.nextInt();
			sumt[i+1]=sumt[i]+t[i];
		}
		for(int i=0; i<n; i++) {
			v[i]=sc.nextInt();
		}

		double maxv[] = new double[sumt[n]*2+1];
		int INF = 1001001;
		for(int i=0; i<maxv.length; i++) {
			maxv[i]=INF;
		}
		int now=0;
		for(int i=0; i<n; i++) {
			for(int j=0; j<2*t[i]; j++) {
				maxv[now] = Math.min(maxv[now], v[i]);
				maxv[now+1] = Math.min(maxv[now+1], v[i]);
				now++;
			}
		}

		maxv[0]=0;
		maxv[sumt[n]*2]=0;

		for(int i=1; i<=sumt[n]*2; i++) {
			maxv[i]=Math.min(maxv[i], maxv[i-1]+0.5);
		}
		for(int i=sumt[n]*2-1; i>=0; i--) {
			maxv[i]=Math.min(maxv[i], maxv[i+1]+0.5);
		}
		double sum = 0.0;
		for(int i=1; i<=sumt[n]*2; i++) {
			sum += (maxv[i] + maxv[i-1]) * 0.25;
		}
		System.out.println(sum);
		sc.close();
	}


}