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;

/**
 * https://abc076.contest.atcoder.jp/tasks/abc076_c
 */
public class Main {

	public static void main(String[] args) throws Exception {
		
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		double[] t = new double[N];
		double[] v = new double[N];
		for(int i=0; i<N; i++) t[i] = sc.nextDouble();
		for(int i=0; i<N; i++) v[i] = sc.nextDouble();
		sc.close();
	
		v[N-1] = Math.min(v[N-1],t[N-1]);
		for(int i=N-2; i>=0; i--){
			v[i] = Math.min(v[i], v[i+1]+t[i]);
		}
		
		double ans = 0;
		double current = 0;
		for(int i=0; i<N; i++){
			double next = i==N-1 ? 0 : v[i]<v[i+1] ? v[i] : v[i+1];
			double maxV = 0;
			double previous = current;
			if(current+t[i]<=next){
				maxV = current+t[i];			
				current = maxV;
			}else{
				maxV = Math.min(v[i], (current+next+t[i])/2);
				ans -= (maxV-next)*(maxV-next)/2;
				current = next;
			}
			ans += maxV*t[i];
			ans -= (maxV-previous)*(maxV-previous)/2;
		}
		System.out.println(ans);
	}

}