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
#include <bits/stdc++.h>
using namespace std;

#define N 100
#define T 200

int t[N];
int limit[N * T * 2];
double acc[N * T * 2 + 1];

int main()
{
	double ans;
	int n;

	cin >> n;
	for( int i = 0; i < n; i++ ) {
		cin >> t[i];
	}
	int total = 0;
	for( int i = 0; i < n; i++ ) {
		int v;
		cin >> v;
		for( int j = total * 2; j < (total + t[i]) * 2; j++ ) {
			limit[j] = v;
		}
		total += t[i];
	}

	acc[0] = 0;
	for( int i = 0; i < total * 2; i++ ) {
		acc[i + 1] = min( acc[i] + 0.5, (double)limit[i] );
	}
	acc[total * 2] = 0;
	for( int i = total * 2 - 1; i >= 0; i-- ) {
		acc[i] = min( acc[i], min( acc[i + 1] + 0.5, (double)limit[i] ) );
	}
	for( int i = 0; i < total * 2; i++ ) {
		ans += (acc[i] + acc[i + 1]) * 0.5 / 2;
	}

	cout << fixed << setprecision(4) << ans << endl;

	return 0;
}
