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 <map>
#include <vector>
#include <iomanip>
using namespace std;

int main() {
	int N; cin >> N;
	vector<int> t(N);
	for(int i=0; i<N; ++i){
	    cin >> t[i];
	    t[i] *= 2;
	}
	map<int, int> diag, diah, diai;
	double sum=0.0;
	for(int i=0; i<N; ++i){
	    int v; cin >> v;
	    diah[sum]=diag[sum]=min(diag[sum], v*2);
	    for(int p=1; p<=t[i]; ++p) diah[sum+p]=diag[sum+p]=v*2;
	    sum += t[i];
	}
	diag[0]=0;
	for(int d=1; d<=sum; ++d) diag[d]=min(diag[d-1]+1, diag[d]);
	diah[sum]=0;
	for(int d=sum-1; d>=0; --d) diah[d]=min(diah[d+1]+1, diah[d]);
	for(int d=0; d<=sum; ++d) diai[d]=min(diag[d], diah[d]);
	double ans=0.0;
	for(int d=1; d<=sum; ++d) ans += (double) (diai[d-1]+diai[d])/8;
	cout << setprecision(15) << ans << endl;
	return 0;
}