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 <bits/stdc++.h>
using namespace std;
using vi=vector<int>;
#define INF 1e9
#define rep(i,n) for(int i=0;i<n;i++)
#define range(i,a,n) for(int i=a;i<n;i++)
#define all(a) a.begin(),a.end()

int main(){
	int n;
	cin>>n;
	vi t(n+1,0), v(n+1,0);
	range(i,1,n+1){cin>>t[i]; t[i]*=2;}
	range(i,1,n+1){cin>>v[i]; v[i]*=2;}
	partial_sum(all(t),t.begin());
	vi m(t[n]+1,INF);
	rep(i,t[n]+1) m[i]=min(m[i],i);
	rep(i,t[n]+1) m[i]=min(m[i],t[n]-i);
	range(i,1,n+1){
		range(j,0,t[i-1]) m[j]=min(m[j],v[i]+t[i-1]-j);
		range(j,t[i-1],t[i]+1) m[j]=min(m[j],v[i]);
		range(j,t[i],t[n]+1) m[j]=min(m[j],v[i]+j-t[i]);
	}
	cout.precision(12);
	cout<<fixed<<accumulate(all(m),0)/4.0<<endl;
	return 0;
}