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) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] times = new int[n + 1];
		int[] velocities = new int[n + 2];
		for (int i = 1; i <= n; i++) {
		    times[i] = sc.nextInt();
		}
		for (int i = 1; i <= n; i++) {
		    velocities[i] = sc.nextInt();
		}
		for (int i = n; i >= 1; i--) {
		    velocities[i] = Math.min(velocities[i], velocities[i + 1] + times[i]);
		}
		for (int i = 1; i <= n; i++) {
		    velocities[i] = Math.min(velocities[i], velocities[i - 1] + times[i]);
		}
		double total = 0;
		for (int i = 1; i <= n; i++) {
		    if (velocities[i - 1] < velocities[i]) {
		        if (velocities[i + 1] < velocities[i]) {
		            total += bothCalc(times[i], velocities[i - 1], velocities[i], velocities[i + 1]);
		        } else {
		            total += leftCalc(times[i], velocities[i - 1], velocities[i]);
		        }
		    } else {
		        if (velocities[i + 1] < velocities[i]) {
		            total += rightCalc(times[i], velocities[i], velocities[i + 1]);
		        } else {
		            total += velocities[i] * times[i];
		        }
		    }
		}
		System.out.println(total);
    }
    
    static double leftCalc(double time, double min, double max) {
        return (2 * time - (max - min)) / 2 * (max - min) + min * time;
    }

    static double rightCalc(double time, double max, double min) {
        return leftCalc(time, min, max);
    }
    
    static double bothCalc(double time, double left, double max, double right) {
        if (left > right) {
            return bothCalc(time, right, max, left);
        }
        if (max - left + max - right <= time) {
            return leftCalc(max - left, left, max) + rightCalc(time - (max - left), max, right);
        }
        return leftCalc(right - left, left, right) + Math.pow(time - (right - left), 2) / 4 + right * (time - (right - left));
    }
    
    
}