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 int long long
const int MOD = 1000000007;
struct edge {int to,costa,costb;};

signed main() {
int n;
cin>>n;
int t[n+1],t0;t[0]=0;
for(int i=0;i<n;i++){
    cin>>t0;
    t[i+1]=t[i]+2*t0;
}
int ans[t[n]+1];
for(int i=0;i<=t[n];i++)ans[i]=min(i,t[n]-i);
priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> que;int v;
for(int i=0;i<n;i++){
    cin>>v;
    que.push(make_pair(2*v,i));
}
pair<int,int>q;
while(!que.empty()){
    q=que.top();que.pop();
    for(int i=t[q.second];i<=t[q.second+1];i++)ans[i]=min(ans[i],q.first);
    for(int i=t[q.second]-1;i>=0&&ans[i]>ans[i+1]+1;i--)ans[i]=ans[i+1]+1;
    for(int i=t[q.second+1]+1;i<=t[n]&&ans[i]>ans[i-1]+1;i++)ans[i]=ans[i-1]+1;
}
float ans2=0;for(int i=0;i<=t[n];i++)ans2+=ans[i];
cout << std::setprecision(20)<<ans2/4;


return 0;


}
