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) {
        new Main().compute();
    }
 
    void compute() {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] t = new int[N + 1];
        int[] v = new int[N + 1];
        for (int i = 0; i < N; i++) {
            t[i] = sc.nextInt();
        }
        for (int i = 0; i < N; i++) {
            v[i] = sc.nextInt();
        }
        for (int i = N - 1; i >= 0; i--) {
            v[i] = Math.min(v[i], v[i + 1] + t[i]);
        }
        double x = 0;
        double vc = 0;
        for (int i = 0; i < N; i++) {
            double vn = v[i + 1];
            if (v[i] < vn) {
                vn = v[i];
            }
            if (vc + t[i] < vn) {
                double vmax = vc + t[i];
                x += vmax * t[i];
                x -= (vmax - vc) * (vmax - vc) / 2;
                vc = vc + t[i];
            } else {
                double vmax = Math.min(v[i], (vc + vn + t[i]) / 2);
                x += vmax * t[i];
                x -= ((vmax - vc) * (vmax - vc) + (vmax - vn) * (vmax - vn)) / 2;
                vc = vn;
            }
        }
        System.out.println(x);
    }
}