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
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <algorithm>
#include <limits>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)

constexpr int MAX_N = 100;
constexpr double INF = 1e15;

int N;
int t[MAX_N + 1];
double v[MAX_N];
double maxv[40005]; // サイズを大きく

int main()
{
	scanf("%d", &N);
	rep(i, N)
		scanf("%d", t + i), t[i] *= 2;
	rep(i, N)
		scanf("%lf", v + i);

	std::fill( maxv, maxv+40005, INF );

	int cur = 0;

	rep(i, N)
	{
		repi(j, cur, cur+t[i]+1)
			maxv[j] = std::min( maxv[j], v[i] ); // ここが肝 (区間の境界のつじつまがあった)

		cur += t[i];
	}

	maxv[0] = 0; maxv[cur] = 0;

	rep(i, cur-1)
		maxv[i + 1] = std::min(maxv[i + 1], maxv[i] + 0.5);

	for (int i = cur - 1; i >= 0; --i)
		maxv[i] = std::min(maxv[i], maxv[i + 1] + 0.5);

	double ans = 0.0;

	rep(i, cur)
		ans += maxv[i] + maxv[i + 1];

	printf("%f\n", ans/4.0);

	return 0;
}