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
#include <cstdio>
#include <iostream>
#include <vector>
using namespace std;

int main(){
  int n;
  cin >> n;
  int t[101];
  int v[102];
  for (int i = 0; i < n; i++){
    scanf("%d", &t[i]);
  }
  for (int i = 0; i < n; i++){
    scanf("%d", &v[i]);
  }
  v[n] = 0;
  float g[60001];
  g[0] = 0;
  int tt = 0;
  for (int i = 0; i < n; i++){
    int u = tt + t[i] * 2;
    for (; tt < u; tt++){
      if (g[tt] + 0.5 <= v[i]) {
        g[tt+1] = g[tt] + 0.5;
      }
      else g[tt+1] = g[tt];
    }
    if (g[tt] > v[i+1]) {
      int p = tt;
      g[p] = v[i+1];
      while (true) {
        if (g[p-1] != g[p] + 0.5){
          g[p-1] = g[p] + 0.5;
          p--;
        } else break;
      }
    }
  }

  float ans = 0;
  for (int i = 0; i < tt; i++){
    ans += (g[i] + g[i+1]);
  }

  printf("%.8f\n", ans / 4);

  return 0;
}
