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;
double f(double x, int *v, int *l, int *r, int N) {
  double V = INT_MAX;
  for (int i = 0; i < N; i++) {
    if (x < l[i]) {
      V = min(V,v[i]+l[i]-x);
    } else if (x <= r[i]) {
      V = min(V,1.0*v[i]);
    } else {
      V = min(V,v[i]+x-r[i]);
    }
  }
  return V;
}
int main() {
  int N;
  cin >> N;
  int t[N+1];
  t[0] = 0;
  for (int i = 1; i <= N; i++) {
    cin >> t[i];
    t[i] += t[i-1];
  }
  N += 2;
  int v[N], l[N], r[N];
  double distance = 0;
  v[0] = 0;
  v[N-1] = 0;
  l[0] = 0;
  r[0] = 0;
  l[N-1] = t[N-2];
  r[N-1] = t[N-2];
  for (int i = 1; i < N-1; i++) {
    l[i] = t[i-1];
    r[i] = t[i];
    cin >> v[i];
  }
  for (double x = 0.5; x <= t[N-2]; x += 0.5) {
    distance += 0.25*(f(x,v,l,r,N)+f(x-0.5,v,l,r,N));
  }
  cout.precision(20);
  cout << fixed << distance;
}