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 <cstdio>
using namespace std;

int main() {
	int n, t[101] = {}, v[101] = {}, i;
	long area, speed = 0, distance4x = 0;
	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d", &t[i]);
	}
	for (i = 0; i < n; i++) {
		scanf("%d", &v[i]);
	}
	for (i = n - 1; i >= 0; i--) {
		if (v[i] - v[i + 1] > t[i]) {
			v[i] = v[i + 1] + t[i];
		}
	}
	for (area = 0; area < n; area++) {
		for (i = 1; i <= t[area]; i++) {
			if (speed > v[area + 1] + (t[area] - i)) {
				speed--;
				distance4x += 2 + 4 * speed;
			}
			else if ((speed + 1 > v[area + 1] + (t[area] - i)) && speed < v[area]) {
				distance4x += 1 + 4 * speed;
			}
			else if (v[area] == speed) {
				distance4x += 4 * speed;
			}
			else {
				speed++;
				distance4x += 4 * speed - 2;
			}
		}
	}
	printf("%f", (double)distance4x / 4.0);
	return 0;
}