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 long long LL;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()

int main() {
  int n; cin >> n;
  vector<double> t(n + 1, 0), v(n + 2, 0);
  LL sum_time = 0;
  for (int i = 1; i <= n; i ++) {
    cin >> t[i];
    sum_time += t[i];
    t[i] = sum_time;
  }
  for (int i = 1; i <= n; i ++) cin >> v[i];

  vector<double> L(n + 2, 0), R(n + 2, 0);
  for (int i = 1; i <= n; i ++) {
    L[i] = R[i - 1];
    R[i] = t[i];
  }
  L[n + 1] = R[n], R[n + 1] = R[n];
  
  double ans = 0;
  double l = 0.0, r;
  for (double i = 0.5; i <= R[n]; i += 0.5) {
    r = 100;
    for (int j = 0; j <= n + 1; j ++) {
      if (0.0 <= i && i <= L[j]) r = min(r, v[j] + L[j] - i);
      else if (L[j] <= i && i <= R[j]) r = min(r, v[j]);
      else r = min(r, v[j] + i - R[j]);
    }
    ans += (l + r) * 0.25;
    l = r;
  }

  cout << setprecision(15) << ans << endl;
}