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 <cstdio>
#define N 105
#define M 40005
using namespace std;

int n, a[M], t[M], v[N];
double s;
int main()
{
	int i, j;
	cin >> n;
	for (i = 1; i <= n; i++) scanf("%d", &t[i]), t[i] *= 2, t[i] += t[i - 1];
	for (i = 1; i <= t[n]; i++) a[i] = 1e9;
	t[n + 1] = t[n];
	for (i = 1; i <= n; i++) scanf("%d", &v[i]), v[i] *= 2;
	a[t[n]] = 0;
	for (i = j = 1; i <= n; i++)
		for (; j <= t[i]; j++) a[j] = min(a[j], min(v[i], a[j - 1] + 1));
	for (i = n + 1, j = t[n] - 1; i > 0; i--)
		for (; j >= t[i - 1]; j--) a[j] = min(a[j], min(v[i], a[j + 1] + 1));
	for (i = 0; i <= t[n]; i++) s += a[i];
	printf("%.6f", s / 4);
    return 0;
}