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>

const int MAX_N = 100;
double t[MAX_N], v[MAX_N];
double l[MAX_N], r[MAX_N];

double f(int index, double time){
   if(time<l[index]) return v[index] + l[index] - time;
   else if(time>r[index]) return v[index] + time - r[index];
   else return v[index];
}

int main(){
   int N;
   std::cin >> N;
   double tsum = 0;
   for(int i=0; i<N; ++i){
      std::cin >> t[i];
      l[i] = tsum;
      tsum += t[i];
      r[i] = tsum;
   }
   for(int i=0; i<N; ++i)
      std::cin >> v[i];

   double ans = 0;
   for(int i=0; i*0.5+0.5<=tsum; ++i){
      double min = 0.25 * (i*0.5 + i*0.5+0.5);
      for(int j=0; j<N; ++j){
	 double area = 0.25 * (f(j, i*0.5) + f(j, i*0.5+0.5));
	 min = std::min(min, area);
      }
      min = std::min(min, 0.25*(r[N-1]-i*0.5 + r[N-1]-i*0.5-0.5));
      ans += min;
   }

   printf("%.4f\n",ans);
   return 0;
}