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.*;

class Main{
	public static void main(String[] $){
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		int[] t=new int[n];
		int[] v=new int[n+1];
		Arrays.setAll(t,i->s.nextInt());
		for(int i=0;i<n;++i)
			v[i]=s.nextInt();
		for(int i=n-1;i>=0;--i)
			v[i]=Math.min(v[i],v[i+1]+t[i]);

		Arrays.parallelPrefix(t,Integer::sum);

		int cv=0,i=0,d=0;
		for(int T=0;T<t[n-1];++T){
			int remain=t[i]-T;
			if(cv-v[i+1]==remain-1&&cv<v[i]){
				d+=cv*4+1;
			}else{
				d+=cv*2;
				if(cv-v[i+1]==remain)
					cv--;
				else if(cv+1<=v[i])
					cv++;
				d+=cv*2;
			}
			if(remain==1)
				++i;
		}
		System.out.println(d/4.0);
	}
}