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;
#define int long long

signed main()
{
   int N; cin>>N;
   vector<double> t(N+1);
   vector<double> v(N+1);
   for(int i=0;i<N;i++) cin>>t[i+1];
   for(int i=0;i<N;i++) cin>>v[i];

   for(int i=0;i<N;i++){
      t[i+1]+=t[i];
   }
   v[N]=0;

   double curT=1;
   double preV=0.0;

   double ans=0.0;
   for(int i=0;i<N;i++){
      while(curT<=2*t[i+1]){
         double curV=preV+0.5;
         curV=min(curV,v[i]);
         for(int j=i+1;j<=N;j++){
            curV=min(curV,v[j]+(2*t[j]-curT)*0.5);
         }
         ans+=(preV+curV)*0.25;
         curT++;
         preV=curV;
      }
   }
   cout<<fixed<<setprecision(10)<<ans<<endl;
}