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:
/*input
1
9
10
*/
#include <bits/stdc++.h>
using namespace std;
const long long md=1e9+7;
int v[105],t[105],n,a[20005],res[20005];

int main(){
	#ifdef ONLINE_JUDGE
	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	#endif
	long long q,m,k,p,x,y;
	cin>>n;
	long double ans=0;
	for(int i=0;i<n;++i){
		cin>>t[i];
	}
	for(int i=0;i<n;++i){
		cin>>v[i];
	}
	int last=0;
	for(int i=0;i<n;last+=t[i++])
		for(int j=0;j<t[i];++j)
			a[last+j]=v[i];

	for(int i=1;i<last;++i)
		res[i]=min({1+res[i-1],a[i],a[i-1]});

	for(int i=last;--i;)
		res[i]=min({res[i],1+res[i+1],a[i]});
	//for(int i=0;i<=last;++i) cout<<i<<" : "<<res[i]<<'\n';
	for(int i=1;i<=last;++i){
		ans+=((double)(res[i-1]+res[i]))/2;
		if(res[i-1]==res[i] && res[i]<a[i-1])
			ans+=0.25;
	}
	cout<<fixed<<setprecision(9)<<ans;
}