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;

int main()
{
	int N, T = 0;
	cin >> N;
	vector<int> t(N);
	for (int i = 0; i < N; i++) {
		cin >> t[i];
		T += t[i];
	}
	vector<int> V(T + 1);
	for (int i = 0, j = 1; i < N; i++) {
		int v;
		cin >> v;
		for (int k = 0; k < t[i]; k++) {
			V[j++] = v;
		}
	}
	vector<vector<double>> dp(T + 1, vector<double>(101, -1e9));
	dp[0][0] = 0;
	for (int i = 1; i <= T; i++) {
		for (int j = 0; j <= V[i]; j++) {
			if (j >= 1) {
				dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + j - 0.5);
			}
			dp[i][j] = max(dp[i][j], dp[i - 1][j] + j + (j < V[i] ? 0.25 : 0.0));
			if (j < V[i]) {
				dp[i][j] = max(dp[i][j], dp[i - 1][j + 1] + j + 0.5);
			}
		}
	}
	cout << fixed << setprecision(3) << dp[T][0] << endl;
	return 0;
}