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;

int N, T[210], V[210];

int t;
int f[40010];
bool memo[40010][210];
double dp[40010][210];

double rec(int idx, double v) {
  if (v < 0) return (-1e20);
  if (v > f[idx]) return (-1e20);
  if (idx == t) return (0.0);
  if (memo[idx][(int)(v * 2)]++) return (dp[idx][(int)(v * 2)]);
  double res = -1e20;
  res = max({res, rec(idx + 1, v) + v * 0.5, rec(idx + 1, v + 0.5) + v * 0.5 + 0.5 * 0.5 / 2,
             rec(idx + 1, v - 0.5) + v * 0.5 - 0.5 * 0.5 / 2});
  dp[idx][(int)(v * 2)] = res;

  return res;
}

int main() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> T[i];
  for (int i = 0; i < N; i++) cin >> V[i];

  t = 1;
  for (int i = 0; i < N; i++) {
    f[t - 1] = min(f[t - 1], V[i]);
    for (int j = 1; j <= 2 * T[i]; j++) f[t++] = V[i];
  }

  f[--t] = 0;
  cout << fixed << setprecision(10) << rec(0, 0) << endl;
}