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, a) for(int i=0; i<(int)(a); i++)
typedef long long ll;
const ll INF = 1LL<<60;

int main()
{
  int n; cin >> n;
  vector<int> t(n);
  int sum = 0;
  rep(i, n) {
    cin >> t[i];
    t[i] *= 2;
    sum += t[i];
  }
  vector<double> v(n);
  rep(i, n) cin >> v[i];
  vector<double> maxv(sum+1, 1001001001.0);
  maxv[0] = 0;
  maxv[sum] = 0;
  int ti = 1;
  rep(i, n) {
    ti--;
    rep(j, t[i]+1) {
      maxv[ti] = min(maxv[ti], v[i]);
      ti++;
    }
  }

  for (int i=1; i<sum; i++) {
    maxv[i] = min(maxv[i], maxv[i-1]+0.5);
  }
  for (int i=sum-1; i>=0; i--) {
    maxv[i] = min(maxv[i], maxv[i+1]+0.5);
  }
  double ans = 0;
  for (int i=1; i<=sum; i++) {
    ans += (maxv[i-1]+maxv[i])/4;
  }
  printf("%.10lf\n", ans);

  return 0;
}