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 <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
#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<double> T(N), V(N), dist(101, -1);
  dist[0] = 0;
  REP(i, N) { cin >> T[i]; }
  REP(i, N) { cin >> V[i]; }
  REP(i, N) {
    REP(j, T[i]) {
      vector<double> ndist(101, -1);
      REP(ps, V[i]+1) {
        for(int s = ps-1; s <= ps+1; ++s) {
          if(s < 0 || s > V[i] || dist[s] < 0) continue;
          double tdist = dist[s] + (s + ps) * 0.5 + (s == ps && s < V[i] ? 0.25 : 0);
          ndist[ps] = max(tdist, ndist[ps]);
        }
      }
      dist = ndist;
    }
  }
  cout << fixed << setprecision(9) << dist[0] << endl;
}