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<stdio.h>
#include<vector>
#include<algorithm>
#include<math.h>
#include<stdlib.h>
#include<string>
#include<iostream>
using namespace std;
int tim[111];
int ve[111];
int maxi[111];
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num; i++)scanf("%d", &tim[i]);
	for (int i = 0; i < num; i++)scanf("%d", &ve[i]);
	int now = 0;
	maxi[0] = 0;
	for (int i = 0; i < num; i++)
	{
		now = min(now + tim[i], ve[i]);
		maxi[i + 1] = now;
	}
	now = 0;
	maxi[num] = 0;
	for (int i = num - 1; i >= 0; i--)
	{
		now = min(now + tim[i], ve[i]);
		maxi[i] = min(maxi[i], now);
	}
	double ans = 0;
	for (int i = 0; i < num; i++)
	{
		double z = (double)(tim[i] - (abs(maxi[i] - maxi[i + 1]))) / 2.0 + max(maxi[i], maxi[i + 1]);
		double m = (maxi[i] + z)*(z - maxi[i]) / 2 + (maxi[i + 1] + z)*(z - maxi[i + 1]) / 2;
		if (z < ve[i])ans += m;
		else ans += m - (z - ve[i])*(z - ve[i]);
	}
	printf("%lf\n", ans);
}