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;

double dp[21000][110];
int T[100];
int V[100];

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
#ifdef LOCAL
	std::ifstream in("in");
	std::cin.rdbuf(in.rdbuf());
#endif

	for(int i = 0; i < 21000; i++){
		for(int j = 0; j < 110; j++){
			dp[i][j] = -1;
		}
	}
	dp[0][0] = 0;

	int N;
	cin >> N;
	for(int i = 0; i < N; i++){
		cin >> T[i];
	}
	for(int i = 0; i < N; i++){
		cin >> V[i];
	}

	int tsum = 0;
	for(int i = 0; i < N; i++){
		int t = T[i], v = V[i];
		for(int j = tsum; j < tsum + t; j++){
			for(int k = 0; k <= v; k++){
				if(dp[j][k] < 0) continue;
				for(int dk = -1; dk <= 1; dk++){
					int nk = k + dk;
					if(nk < 0 || v < nk) continue;
					double sum = dp[j][k] + (k + nk) / 2.0;
					if(dk == 0 && k < v) sum += 0.25;
					dp[j + 1][nk] = max(dp[j + 1][nk], sum);
				}
			}
		}
		tsum += t;
	}
	cout << fixed << setprecision(15) << dp[tsum][0] << endl;
}