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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

int main() {
  int n;
  cin >> n;
  vector<int> ts(n + 2), vs(n + 2);
  rep(i, n) cin >> ts[i + 1];
  rep(i, n) cin >> vs[i + 1];
  for (int i = 1; i <= n; i++) {
    vs[i] = min(vs[i], vs[i - 1] + ts[i]);
  }
  for (int i = n; i >= 1; i--) {
    vs[i] = min(vs[i], vs[i + 1] + ts[i]);
  }
  double ans = 0;
  for (int i = 1; i <= n; i++) {
    double t = ts[i], d1 = vs[i] - vs[i - 1], d2 = vs[i] - vs[i + 1];
    if (d1 > 0 && d2 > 0 && d1 + d2 > t) {
      double vl = min(vs[i - 1], vs[i + 1]), vd = vl + t - max(vs[i - 1], vs[i + 1]);
      ans += (vl * 2 + t) * t / 2 - vd * vd / 4;
    } else {
      ans += vs[i] * ts[i];
      if (d1 > 0) ans -= d1 * d1 / 2;
      if (d2 > 0) ans -= d2 * d2 / 2;
    }
  }
  cout << setprecision(20) << ans << endl;
  return 0;
}