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:
// practice with Dukkha
#include <algorithm>
#include <iostream>

using namespace std;

const int N = 100, S = 20000;

int tt[N], vv[N], oo[S], pp[S], qq[S];

int main() {
	int n; cin >> n;
	int s = 0;
	for (int i = 0; i < n; i++) {
		int t; cin >> t;
		s += tt[i] = t;
	}
	for (int i = 0; i < n; i++) {
		int v; cin >> v;
		vv[i] = v;
	}
	for (int h = 0; h < s; h++) {
		oo[h] = s;
		pp[h] = h;
		qq[h] = s - 1 - h;
	}
	for (int l = 0, i = 0; i < n; i++) {
		int t = tt[i], v = vv[i];
		int r = l + t;
		for (int h = l; h < r; h++)
			oo[h] = min(oo[h], v);
		for (int h = r; h < s; h++)
			pp[h] = min(pp[h], v + h - r);
		for (int h = 0; h < l; h++)
			qq[h] = min(qq[h], v + l - 1 - h);
		l = r;
	}
	double ans = 0;
	for (int h = 0; h < s; h++) {
		int o = oo[h], p = pp[h], q = qq[h];
		if (o <= p && o <= q)
			ans += o;
		else
			ans += p == q ? p + 0.25 : min(p, q) + 0.5;
	}
	printf("%f\n", ans);
	return 0;
}