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;

using ll = long long;


int main() {
	int N;
	cin >> N;
	vector<int> t(N), v(N);
	for (int i = 0; i < N; i++) {
		cin >> t[i];
		t[i] *= 2;
		if (i) t[i] += t[i-1];
	}
	for (int i = 0; i < N; i++) {
		cin >> v[i];
		v[i] *= 2;
	}

	vector<int> res(t.back() + 1);
	for (int i = 0; i < res.size(); i++) {
		res[i] = t.back() - i;
	}
	for (int i = 0; i < N - 1; i++) {
		for (int j = t[i]; j; j--) {
			res[j] = min(res[j], t[i] + v[i+1] - j);
		}
	}
	for (int i = 0; i < N; i++) {
		int s = i ? t[i-1] : 0;
		for (int j = s; j <= t[i]; j++) {
			res[j] = min(res[j], v[i]);
		}
	}
	res[0] = 0;
	double r = 0;
	for (int i = 1; i < res.size(); i++) {
		res[i] = min(res[i], res[i-1] + 1);
		r += res[i];
	}
	cout << fixed << setprecision(12) << r / 4 << endl;
}