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>
#define LL long long
using namespace std;

double a[100010];
double v[100010];


int main()
{
	
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++)
		scanf("%lf", a + i);
	for(int i = 1; i <= n; i++)
		scanf("%lf", v + i);
	v[n + 1] = 0;
	double ans = 0;
	for(int i = n; i >= 1; i--)
	{
		if(v[i] - a[i] > v[i + 1])
			v[i] = v[i + 1] + a[i];
	}
	double x = 0;
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= a[i]; j++)
		{
			int dt = a[i] - j;
			
			if(x - dt > v[i + 1])
				x--, ans += 0.5 + x;
			else if(x + 1 - dt > v[i + 1] && x < v[i])
				ans += 0.25 + x;
			else if(x == v[i])
				ans += x;
			else 
				ans += x + 0.5, x++; 
		}		
	}	
	printf("%.10lf", ans);
	return 0;
}