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 < N; i++)
const int INF = INT_MAX / 2;

int main() {
  int N;
  cin >> N;
  vector<int> t(N), v(N);
  rep(i, N) cin >> t[i];
  rep(i, N) cin >> v[i];
  int T = 0;
  rep(i, N) T += t[i];
  vector<int> ub(T * 2 + 1, INF);
  int x = 0;
  rep(i, N) rep(y, t[i]) {
    ub[x] = min(ub[x], v[i] * 2);
    ub[x + 1] = min(ub[x + 1], v[i] * 2);
    ub[x + 2] = min(ub[x + 2], v[i] * 2);
    x += 2;
  }
  ub[0] = ub[T * 2] = 0;
  rep(x, T * 2) ub[x + 1] = min(ub[x + 1], ub[x] + 1);
  for (int x = T * 2; x > 0; x--) ub[x - 1] = min(ub[x - 1], ub[x] + 1);
  double ans = 0;
  rep(x, T * 2) ans += (ub[x] + ub[x + 1]) / 2.0;
  cout << fixed << setprecision(12) << ans / 4 << endl;
}