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;
typedef long long ll;
ll dp[40005][205]={0};
ll max_v[40005]={0};
ll ts[205]={0};
const ll inf=174417441744;
int main(){
  ll n;
  cin>>n;
  ll all_time=0;
  for(int i=0;i<n;i++) cin>>ts[i];
  for(int i=0;i<n;i++){
    ll v,t=ts[i];
    cin>>v;
    v*=2;
    for(int j=all_time+1;j<=all_time+t*2;j++) max_v[j]=v;
    all_time+=t*2;
  }
  for(int i=0;i<=all_time;i++){
    for(int j=0;j<205;j++) dp[i][j]=-inf;
  }
  dp[0][0]=0;
  for(int i=0;i<all_time;i++){
    for(int j=0;j<=max_v[i+1];j++){
      if(j>0) dp[i+1][j-1]=max(dp[i+1][j-1],dp[i][j]+j+j-1);
      dp[i+1][j]=max(dp[i+1][j],dp[i][j]+2*j);
      if(j+1<=max_v[i+1]) dp[i+1][j+1]=max(dp[i+1][j+1],dp[i][j]+j+j+1);
    }
  }
  cout<<fixed;
  cout<<setprecision(5)<<(double)dp[all_time][0]/8<<endl;
}