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;
const int maxn=105;

int n;
double v[maxn],t[maxn];
double cur,ans;

int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%lf",&t[i]);
	}
	for(int i=0;i<n;i++){
		scanf("%lf",&v[i]);
	}
	for(int i=0;i<n;i++){
		double inc,kep,dec;
		double lim=100000000.0;
		double tmp=0.0;
		for(int j=i+1;j<=n;j++){
			lim=min(lim,v[j]+tmp);
			tmp+=t[j];
		}
		inc=min(v[i]-cur,(lim+t[i]-cur)*0.5);
        inc=min(inc,t[i]);
        dec=max(0.0,cur+inc-lim);
        kep=t[i]-inc-dec;
        ans+=0.5*(cur*2.0+inc)*inc;
        ans+=(cur+inc)*kep;
        ans+=0.5*((cur+inc)*2.0-dec)*dec;
        cur+=inc;
        cur-=dec;
	}
	printf("%f",ans);
	return 0;
}