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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)

int main(){
  int N;
  cin >> N;
  double t, v[N+1], ta[N];
  rep(i, N) {
    cin >> t;
    if(!i) ta[i] = t;
    else ta[i] = ta[i-1] + t;
  }
  rep(i, N) cin >> v[i];
  v[N] = 0.0, t = 0;
  int pos = 0;
  double ans = 0, vel = 0.0;
  while(t < ta[N-1]){
    bool can1 = true, can0 = true;
    for(int i = pos; i < N; i++){
      if(ta[i] - t  == vel - v[i+1]) can0 = false;
      if(ta[i] - t  - (vel - v[i+1]) == 1.0) can1 = false;
    }
    if(!can0) {
      ans += vel - 0.5;
      vel--;
    }
    else if(can0 && !can1){
      if(vel == v[pos]) ans += vel;
      else ans += vel + 0.25;
    }
    else if(vel < v[pos]){
      ans += vel + 0.5;
      vel++;
    } else ans += vel;
    t++;
    if(t >= ta[pos]) pos++;
  }
  printf("%6f\n", ans);
  return 0;
}