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;

//https://atcoder.jp/contests/abc076/tasks/abc076_d

int n, x[103], v[102], m;
double l, r, ans = 0;

double f(double t, int i) {
	if (t < x[i])
		return v[i] + x[i] - t;
	if (t > x[i + 1])
		return v[i] + t - x[i + 1];
	return v[i];
}

double sup(double t) {
	double rv = m;
	for (int i = 0; i < n + 2; i++) {
		double s = f(t, i);
		rv = min(rv, s);
	}
	return rv;
}

int main() {
	cin >> n;
	for (int i = 0, t; i < n; i++) {
		cin >> t;
		x[i + 2] = x[i + 1] + t;
	}
	x[n + 2] = x[n + 1];
	for (int i = 0; i < n; i++) {
		cin >> v[i + 1];
		m = max(m, v[i + 1]);
	}
	for (double t = 0; t < x[n + 1]; t += 0.5)
		ans += sup(t) + sup(t + 0.5);
	cout <<fixed<< ans / 4 << endl;
}