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 <stdio.h>

int sum1to2(int* a,int x,int y){
  int ans=0;
  for(int i=x;i<=y;i++){
    ans+=a[i];
  }
  return ans;
}
int main(){
  int N;
  scanf("%d",&N);
  int lt[N];
  int lv[N+1];
  for(int i=0;i<N;i++){
    scanf("%d",&(lt[i]));
  }
  for(int i=0;i<N;i++){
    scanf("%d",&(lv[i]));
  }
  lv[N]=0;
  double x=0;
  double v=0;
  double pv=0;
  double t;
  double dt=0.5;
  int flag=0;
  for(int i=0;i<N;i++){
    t=0;
    	if(v>lv[i]){
	  v=(double)lv[i];
	  }
    while(t<lt[i]){
      flag=0;
      pv=v;
      for(int j=i;j<N;j++){
	if(sum1to2(lt,i,j)-t<=v-lv[j+1]){
	  flag=1;
	}
      }
      if(flag==1){//残り時間で下げられる速度が下げなきゃいけない速度より小さい
	v-=1*dt;
      }else if(v<lv[i]){
	v+=1*dt;
	if(v>lv[i]){
	  v=(double)lv[i];
	  }
      }
      x+=(v+pv)/2.0*dt;
      t+=dt;
    }
  }
  printf("%f",x);
  return 0;
}