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;

const int INF = 1000000000;

int main()
{
  int N; scanf("%d", &N);
  vector<int> T(N + 1, 0), V(N + 1, 0);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &T[i]);
    T[i] *= 2;
    T[i] += T[i - 1];
  }

  for (int i = 1; i <= N; ++i) {
    scanf("%d", &V[i]);
  }

  vector<double> upperV(T[N] + 5, 0);
  double limv = V[1];
  for (int t = 1, vpos = 0; t <= T[N] ; ++t) {
    if (t > T[vpos]) { limv = V[++vpos]; }
    upperV[t] = min(upperV[t - 1] + 0.5, limv);
  }

  upperV[T[N]] = 0;
  limv = V[N];
  for (int t = T[N] - 1, vpos = N; t >= 0; --t) {
    if (t < T[vpos - 1]) { limv = V[--vpos]; }
    upperV[t] = min(upperV[t], min(upperV[t + 1] + 0.5, limv));
  }

  double ans = 0;
  for (int t = 1; t <= T[N]; ++t) {
    ans += (upperV[t - 1] + upperV[t]);
  }
  ans *= 0.25;
  printf("%f\n", ans);
  return 0;
}