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 <string.h>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;
typedef long long ll;
const int INF=100000000LL;
int main() {
	int n;
	double t[102],v[102];
	cin>>n;
	for(int i=1;i<=n;++i){
		cin>>t[i];
		t[i]+=t[i-1];
	}
	for(int i=1;i<=n;++i)cin>>v[i];
	t[n+1]=t[n];
	t[n+2]=t[n];
	v[n+1]=0;
	double a[40000];
	a[0]=0;
	double ans=0;
	for(int i=1;i<=t[n+1]*2;++i){
		double k=(double)i/2;
		a[i]=min(k,t[n+1]-k);
		for(int j=1;j<=n;++j){
			double l=t[j-1],r=t[j];
			if(k<l)a[i]=min(a[i],v[j]+(l-k));
			if(k>r)a[i]=min(a[i],v[j]+(k-r));
			if(k>=l&&k<=r)a[i]=min(a[i],v[j]);

		}
//		cout<<i<<" "<<a[i]<<endl;
		ans+=(a[i-1]+a[i])/4;
	}
	printf("%f\n",ans);
	return 0;
}