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;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; i++)
#define all(a) (a).begin(), (a).end()

int N;
vector<int> t, v;

int main () {
  cin >> N;
  t.resize(N);
  v.resize(N+1, 0);
  rep(i,N) {
    cin >> t[i];
    t[i] *= 2;
  }
  rep(i,N) {
    cin >> v[i];
    v[i] *= 2;
  }

  vector<int> d(N+1, 0);
  rep(i,N) d[i+1] = d[i] + t[i];
  rep(i,N+1) d[i] += v[i];
  for(int i = N - 1; i >= 0; i--) {
    d[i] = min(d[i], d[i+1]);
  }

  double x = 0;
  int vc = 0;
  int tc = 0;
  rep(i,N) {
    int tb = tc;
    rep(j,t[i]) {
      if(vc >= - tc + d[i+1]) {
        x += vc - 0.5;
        vc -= 1;
      } else if(vc >= v[i]) {
        x += vc;
      } else {
        x += vc + 0.5;
        vc += 1;
      }
      tc += 1;
    }
  }

  printf("%.4lf\n", x*0.25);
  return 0;
}