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<iostream>
#include<algorithm>
#include<vector>
#include<tuple>
#define TT std::tuple<double,double,double>
double t[210];
double v[210];
std::vector<TT> D;
double ans;
int SE;
int main() {
	int N;
	std::cin >> N;
	D.push_back(TT{ 0,0,0 });
	for (int i = 1; i <= N; i++) {
		std::cin >> t[i];
		t[i] += t[i - 1];
	}
	for (int i = 1; i <= N; i++) {
		std::cin >> v[i];
		D.push_back(TT{ t[i - 1],t[i],v[i] });
	}
	D.push_back(TT{ t[N],t[N],0 });
	double bef = 0;
	for (double i = 0.5; i <= t[N]; i+=0.5) {
		double ms = bef + 0.5;
		for (auto u : D) {
			double a, b, c;
			std::tie(a, b, c) = u;
			if (b <= i) {
				ms = std::min(ms, c + i - b);
			}
			if (a <= i&&i <= b) {
				ms = std::min(ms, c);
			}
			if (i <= a) {
				ms = std::min(ms, c + a - i);
			}
		}
		ans += (bef + ms)*0.25;

		bef = ms;
	}
	std::cout << std::fixed << ans << std::endl;
	return 0;
}