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 "bits/stdc++.h"
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
typedef pair<P, int> PP;
int t[100];
int v[100];
int main() {
	int N;
	cin >> N;
	vector<PP> vec;
	for (int i = 0; i < N; i++) cin >> t[i];
	for (int i = 0; i < N; i++) cin >> v[i];
	int T = 0;
	vec.push_back(PP(P(0, 0), 0));
	for (int i = 0; i < N; i++) {
		vec.push_back(PP(P(T, T + t[i]), v[i]));
		T += t[i];
	}
	vec.push_back(PP(P(T, T), 0));
	double p_max_v = 0;
	double ans = 0;
	for (int i = 1; i <= 2 * T; i++) {
		double max_v = 1e9;
		double t = i / 2.0;
		for (int j = 0; j < vec.size(); j++) {
			if (t <= vec[j].first.first) max_v = min(max_v, vec[j].second + (vec[j].first.first - t));
			else if (t >= vec[j].first.second) max_v = min(max_v, vec[j].second + (t - vec[j].first.second));
			else max_v = min(max_v, (double)vec[j].second);
		}
		ans += 0.25*(max_v + p_max_v);
		p_max_v = max_v;
	}
	printf("%.15lf\n", ans);
}