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>
#define INF 2147483647
#define INF_LL 9223372036854775807
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;

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

	vector<double> A;
	for (double a = 0; a <= t_s[N]; a += 0.5) {
		vector<double> v_f;
		v_f.push_back(a);
		v_f.push_back(t_s[N] - a);
		for (int i = 0; i < N; i++) {
			if (a <= t_s[i])v_f.push_back(v[i] + (t_s[i] - a));
			else if (t_s[i] < a && a <= t_s[i + 1])v_f.push_back(v[i]);
			else if (t_s[i + 1] < a)v_f.push_back(v[i] + (a - t_s[i + 1]));
		}
		A.push_back(*min_element(v_f.begin(), v_f.end()));
	}

	double ans = 0;
	for (int i = 0; i < A.size() - 1; i++) {
		ans += 0.25*(A[i] + A[i + 1]);
	}
	cout << setprecision(8) << ans << endl;
	return 0;
}