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;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] t = new int[n];
		int[] v = new int[n+1];
		for(int i=0;i<n;i++) {
			t[i] = in.nextInt();
		}
		for(int i=0;i<n;i++) {
			v[i] = in.nextInt();
		}
		int limitSpeed = 0;
		for(int i=n-1;i>=0;i--) {
			if(v[i]>limitSpeed+t[i]) {
				v[i] = limitSpeed+t[i];
				limitSpeed = limitSpeed+t[i];
			}else {
				limitSpeed = v[i];
			}
		}
		int maxSpeed = 0;
		for(int i=0;i<n;i++) {
			if(v[i]>maxSpeed+t[i]) {
				v[i] = maxSpeed+t[i];
				maxSpeed = maxSpeed+t[i];
			}else {
				maxSpeed = v[i];
			}
		}
		double ans = 0;
		int vnow = 0;
		for(int i=0;i<n;i++) {
			if(v[i]>=vnow+t[i] && (v[i+1]>=vnow+t[i])) {
				ans += (double)vnow*t[i] + (double)t[i]*t[i]/2;
				vnow += t[i];
			}else if(v[i]>=((double)t[i]+vnow+v[i+1])/2) {
				ans += ((double)t[i]+vnow+v[i+1])*t[i]/2 - ((double)t[i]-vnow+v[i+1])*((double)t[i]-vnow+v[i+1])/8 - ((double)t[i]+vnow-v[i+1])*((double)t[i]+vnow-v[i+1])/8;
				vnow = v[i+1];
			}else {
				ans += v[i]*t[i] - ((double)v[i]-vnow)*((double)v[i]-vnow)/2 - (v[i]>v[i+1]?((double)v[i]-v[i+1])*((double)v[i]-v[i+1])/2:0);
				vnow = Math.min(v[i], v[i+1]);
			}
		}
		System.out.println(ans);
		in.close();

	}

}