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>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <stdio.h>
#include <queue>
using namespace std;
#define rep(i,n) for(i=0;i<n;i++)
typedef long long ll;
typedef pair <int,ll> pii;

double trap(double h1,double h2,double w)
{
  return 0.5*(h1+h2)*w;
}

int main(void)
{
  int n,tt;
  cin >> n;
  vector<int> t(n+5,0);
  cin >> t[0];
  vector<int> v(n+5,0);
  for(int i=1;i<n;i++)
    {
      cin >> tt;
      t[i]=tt+t[i-1];
    }
  for(int i=0;i<n;i++)
    {
      cin >> v[i];
    }
  v[n]=0;
  double cv=0.0,ans=0.0,prev=0.0;
  int ct=0;
  for(double i=0.5;i<=t[n-1];i+=0.5)
    {
      cv+=0.5;
      if(i>t[ct]) ct++;
      for(int j=0;j<n;j++)
	{
	  if(i<=t[j]) cv=min(cv,v[j+1]+(t[j]-i));
	}
      cv=min(cv,(double)v[ct]);
      ans+=trap(prev,cv,0.5);
      prev=cv;
    }
  cout  << fixed << ans << endl;
  return 0;
}