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 rep(i,n) for(int i=0;i<n;i++)

int main() {
  int n;
  cin>>n;
  int t[n],v[n];
  rep(i,n) cin>>t[i];
  rep(i,n) cin>>v[i];
  int s[n+1];
  s[0]=0;
  rep(i,n){
    s[i+1]=s[i]+t[i];
  }
  double ans[2*s[n]+1];
  rep(i,2*s[n]+1){
    ans[i]=min(i/2.0,s[n]-i/2.0);
  }
  rep(i,2*s[n]+1){
    rep(j,n){
      if(i/2.0<s[j]){
        ans[i]=min(ans[i],v[j]+s[j]-i/2.0);
      }
      else if(i/2.0>s[j+1]){
        ans[i]=min(ans[i],v[j]-s[j+1]+i/2.0);
      }
      else{
        ans[i]=min(ans[i],v[j]+0.0);
      }
    }
  }
  double res=0;
  rep(i,2*s[n]){
    res+=(ans[i]+ans[i+1])/4.0;
  }
  cout<<fixed<<setprecision(5);
  cout<<res<<endl;
}