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;
int t[202];
int v[202];
const double INF = 10000000.0;

int main() {
  cin >> n;

  for(int i = 0; i < n; i++) {
    cin >> t[i];
  }
  for(int i = 0; i < n ;i++) {
    cin >> v[i];
  }

  vector<double> dp(101,-INF);
  dp[0] = 0.0;

  for(int i = 0; i < n; i++) {
    for(int j = 0; j < t[i]; j++) {
      vector<double> nextdp(101,-INF);
      for(double k = 0; k <= v[i]; k++) {
	nextdp[k] = max(nextdp[k],dp[k] + k);
	if(k < v[i]) {
	  nextdp[k] = max(nextdp[k],dp[k] + k + 0.25);
	  nextdp[k+1] = max(nextdp[k+1],dp[k] + k + 0.5);
	}
	if(k > 0)
	  nextdp[k-1] = max(nextdp[k-1],dp[k] + k - 0.5);
      }
      dp.swap(nextdp);
    }
  }

  cout << fixed << dp[0] << endl;
}