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>
#define rep(i, n) for(int i=0;i<n;i++)
#define repb(i, a, b) for(int i=a;i<b;i++)

using namespace std;
typedef pair<int, long> P;
typedef long long int ll;

const int INF = 1<<30;

int main(){
  int n;cin>>n;
  vector<int> t(n+1), v(n);t[0]=0;
  rep(i, n)cin>>t[i+1];
  rep(i, n){
  	t[i+1]+=t[i];
  }
  rep(i, n)cin>>v[i];
  int Mt=t[n];
  int dtime=1;
  int dmv, pdmv=0;
  int ans=0;
  while(dtime<=Mt*2){
    dmv=min(dtime, Mt*2-dtime);
    rep(i, n){
      if(t[i]*2<=dtime && dtime<=t[i+1]*2){
      	dmv=min(dmv, v[i]*2);
      }else if(dtime<t[i]*2){
        dmv=min(dmv, (t[i]*2-dtime)+v[i]*2);
      }else{
      	dmv=min(dmv, (dtime-t[i+1]*2)+v[i]*2);
      }
    }
    // cout<<dtime*0.5<<"sec: "<<dmv*0.5<<"m/s"<<endl;
    ans+=(dmv+pdmv);
    pdmv=dmv;
    dtime++;
  }
  cout<<fixed<<setprecision(10)<<ans*0.125<<endl;
  return 0;
}