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<iostream>
using namespace std;
typedef long long int ll;
typedef long double ld;
int main(){
  ll n,t[110],v[110],ts[110]={};
  ld a[50000]={},ans=0;
  cin>>n;
  ts[0]=0;
  for(int i=0;i<n;i++){cin>>t[i];ts[i+1]=ts[i]+t[i];}
  for(int i=0;i<n;i++) cin>>v[i];
  for(ll i=0;i<=2*ts[n];i++){
    a[i]=  min((ld)i/2,(ld)ts[n]-(ld)i/2);
    for(int j=0;j<n;j++){
      if(i < 2*ts[j]) a[i]=  min(a[i],(ld)v[j]+(ld)ts[j]-(ld)i/2);
      else if( i > 2*ts[j+1]) a[i]= min(a[i],(ld) v[j]+(ld)i/2-(ld)ts[j+1]);
      else a[i]=min(a[i],(ld) v[j]);
    }
  }
  for(int i=0;i<2*ts[n];i++){
    ans+=(a[i]+a[i+1])/4;
  }
  printf("%f",(double)ans);
}