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;

const int inf = 1e9;

int n;
vector<int> t, v;

//i is parameter(0-n+1)
int fni(int i, int tm) {
  int ret;
  if (i == 0)
    ret = tm;
  else if (i == n + 1)
    ret = t[n] - tm;
  else if (tm < t[i - 1])
    ret = -(tm - t[i - 1]) + v[i];
  else if (tm > t[i])
    ret = (tm - t[i]) + v[i];
  else
    ret = v[i];
  return ret;
}

int fn(int tm) {
  int ret = fni(0, tm);
  for (int i = 1; i <= n + 1; i++) {
    ret = min(ret, fni(i, tm));
  }
  return ret;
}

int main() {
  cin >> n;
  t.resize(n + 1, 0);
  v.resize(n + 1, 0);

  for (int i = 1; i <= n; i++) {
    cin >> t[i];
    t[i] *= 2;
  }
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    v[i] *= 2;
  }


  for (int i = 0; i < n; i++) t[i + 1] += t[i];


  double ans = 0;
  for (int i = 0; i < t[n]; i ++) {
    ans += (fn(i) + fn(i + 1));

  }
    cout << fixed << setprecision(10) << ans / 8.0 << endl;

}