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;
import java.util.stream.IntStream;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        int[] t = readArray(sc, N);
        int[] v = readArray(sc, N);

        System.out.println(solve(t, v));

        sc.close();
    }

    static int[] readArray(Scanner sc, int size) {
        int[] result = new int[size];
        for (int i = 0; i < result.length; ++i) {
            result[i] = sc.nextInt();
        }

        return result;
    }

    static double solve(int[] t, int[] v) {
        int N = t.length;

        int leftMax = 0;
        int[] leftMaxs = new int[N - 1];
        for (int i = 0; i < leftMaxs.length; ++i) {
            leftMax = Math.min(Math.min(leftMax + t[i], v[i]), v[i + 1]);
            leftMaxs[i] = leftMax;
        }

        int rightMax = 0;
        int[] rightMaxs = new int[N - 1];
        for (int i = rightMaxs.length - 1; i >= 0; --i) {
            rightMax = Math.min(Math.min(rightMax + t[i + 1], v[i + 1]), v[i]);
            rightMaxs[i] = rightMax;
        }

        int[] maxs = IntStream.range(0, N - 1).map(i -> Math.min(leftMaxs[i], rightMaxs[i])).toArray();

        return IntStream.range(0, N)
                .mapToDouble(
                        i -> computeMaxDistance(t[i], v[i], (i == 0) ? 0 : maxs[i - 1], (i == N - 1) ? 0 : maxs[i]))
                .sum();
    }

    static double computeMaxDistance(int time, int limitV, int v1, int v2) {
        double maxV;
        if ((limitV - v1) + (limitV - v2) <= time) {
            maxV = limitV;
        } else {
            maxV = (time + v1 + v2) / 2.0;
        }

        return (v1 + maxV) * (maxV - v1) / 2 + maxV * (time - (maxV - v1) - (maxV - v2))
                + (maxV + v2) * (maxV - v2) / 2;
    }
}