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 + 2], v = new int[n + 2];
        int tSum = 0;
        for (int i = 1; i <= n; i++) {
            t[i] = in.nextInt() * 2;
            tSum += t[i];
        }
        for (int i = 1; i <= n; i++) {
            v[i] = in.nextInt();
        }
        double[] vMax = new double[tSum + 1];
        int idx = 1;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < t[i]; j++) {
                vMax[idx + j] = v[i];
            }
            idx += t[i];
        }
//        System.out.println(Arrays.toString(vMax));
        idx = 0;
        for (int i = 0; i < n; i++) {
            double max = v[i];
            int tmp = idx + 1;
            while(max < vMax[tmp]) {
                max += 0.5;
                vMax[tmp++] = max;
                if(vMax.length <= tmp) break;
            }
            idx += t[i+1];
        }
        idx = tSum + 1;
        for (int i = n+1; i > 0 ; i--) {
            double max = v[i];
            int tmp = idx - 1;
            while(max < vMax[tmp]) {
                vMax[tmp--] = max;
                max += 0.5;
                if(tmp < 0) break;
            }
            idx -= t[i-1];
        }
//        for (int i = 0; i < vMax.length; i++) {
//            System.out.println("idx = " + i  + " time = " + (i / 2.0) + " v = " + vMax[i]);
//        }
        double ans = 0, now = 0;
        for (int i = 0; i < vMax.length - 1; i++) {
            ans += now * 0.5;
            if(vMax[i] < vMax[i + 1]) {
                ans += 0.125;
                now += 0.5;
            }
            if(vMax[i + 1] < vMax[i]) {
                ans -= 0.125;
                now -= 0.5;
            }
        }
        System.out.println(ans);
    }
}