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 CPP):

Solution:
#include <iostream>
#include <cstdio>
using namespace std;

int n, num,ti[105], vi[105];
double mx, s, Y, d[105][105];
int f(int p) {return p > 0 ? p : -p;}
int main()
{
	int i, j, k;
	cin >> n;
	for (i = 1; i <= n; i++) {
		scanf ("%d", &ti[i]);
	}
	for (i = 1; i <= n; i++) {
		scanf ("%d", &vi[i]);
	}
	for (i = 1; i <= n; i++) {
		for (j = 0; j <= vi[i]; j++) {
			mx = -2e9;
			for (k = 0; k <= vi[i - 1]; k++) {
				if (k > vi[i]) continue;
				if (f(j - k) > ti[i]) continue;
				num = vi[i] - k + vi[i] - j;
				if (num <= ti[i]) s =vi[i] * ti[i] - (double)(vi[i] - k) * (vi[i] - k) / 2 - (double)(vi[i] - j) * (vi[i] - j) / 2;
				else {
					Y = (double)(ti[i] + j + k) / 2;
					s = Y * ti[i] - (double)(Y - k) * (Y - k) / 2 - (double)(Y - j) * (Y - j) / 2;
				}
				s += d[i - 1][k];
				if (mx == -2e9 || s > mx) mx = s;
			}
			d[i][j] = mx;
		}
	}
	printf("%.7f", d[n][0]);
    return 0;
}