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;

int main(){
  int n;
  double ans = 0.0, nowv = 0.0, nowt = 0.0, lim = 0.0;
  cin >> n ;
  int t[n+1], sumt=0;
  int cnt = 1;
  t[0] = 0;
  int v[n+2];
  v[n+1] = 0;
  v[0] = 0;
  for(int i=1;i<=n;i++){
    cin >> t[i];
    sumt += t[i];
    t[i] = sumt;
  }
  for(int i=1;i<=n;i++){
    cin >> v[i];
  }
  for(double i = 0.5; i<=sumt;i+=0.5){
    if(i > t[cnt]) cnt++;
    double lim = v[cnt];
    double tmp;
    for(int j=0;j<=n;j++){
      if(j < cnt){
        tmp = v[j] + i - t[j];
      }else{
        tmp = v[j+1] + t[j] - i;
      }
      lim = min(lim, tmp);
    }
    if(nowv<lim){
      ans += (nowv/2) + 0.25;
      nowv += 0.5;
    }else if(nowv>lim){
      ans += (nowv/2) - 0.25;
      nowv -= 0.5;
    }else{
      ans += (nowv/2);
    }
  }
  cout << fixed << setprecision(8) <<  ans << endl;
}