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 <stdio.h>

using namespace std;

int N;
int t[205], v[205];
int upper[40005];
int T;


int main(void)
{
	cin >> N;
	for(int i = 0; i < N; i++){
		cin >> t[i];
		t[i] *= 2;
	}
	for(int i = 0; i < N; i++){
		cin >> v[i];
		v[i] *= 2;
	}
	for(int i = 0; i < N; i++) T += t[i];
	
	for(int i = 0; i <= T; i++) upper[i] = min(i, T-i);
	int lb = 0;
	for(int i = 0; i < N; i++){
		for(int j = lb; j <= lb + t[i]; j++) upper[j] = min(upper[j], v[i]);
		for(int j = lb; j >= 0; j--) upper[j] = min(upper[j], v[i] + (lb-j));
		lb += t[i];
		for(int j = lb; j <= T; j++) upper[j] = min(upper[j], v[i] + (j-lb));
	}
	
	int ans = 0;
	for(int i = 0; i < T; i++){
		ans += upper[i] * 2;
		ans += upper[i] - upper[i+1];
	}
	printf("%.8f\n", ans / 8.0);
	return 0;
}