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 unsigned long long ul;
typedef signed long long ll;
ul over = 1000000007;

int main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed;
  int n;
  cin >> n;
  int t[n+2], v[n+2];
  t[0] = 0; t[n+1] = 0; v[0] = 0; v[n+1] = 0;
  for (int i=0; i<n; ++i) cin >> t[i+1];
  for (int i=0; i<n; ++i) cin >> v[i+1];
  double ans = 0, speed = 0;
  for (int i=1; i<=n; ++i) {
    for (int j=0; j<t[i]*2; ++j) {
      bool flg = false;
      int ov_time = 0;
      for (int k=i+1; k<=n+1; ++k) {
        if (speed - (t[i]*2-j-1)/2.0-ov_time > v[k]) {flg = true; break;}
        ov_time += t[k];
      }
      if (flg) {
        ans += (speed+speed-0.5)*0.5/2;
        speed -= 0.5;
      } else {
        if ((int)speed == v[i]) ans += speed * 0.5;
        else {
          ans += (speed+speed+0.5)*0.5/2;
          speed += 0.5;
        }
      }
    }
  }
  cout << ans << endl;
  return 0;
}