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;
typedef long long ll;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
const int N=409;
const int T=80010;
int v[N],t[N],Vg[T],n,Vw[T];
double ans;
 
int main(){
	cin >> n;
	CLR(Vw);CLR(Vg);CLR(v);CLR(t);
	FOR(i,0,n){
		cin >> t[i];
	}
	FOR(i,0,n){
		cin >> v[i];
	}
	int tw=0;
	Vg[0]=0;++tw;
	FOR(i,0,n){
		FOR(j,0,4*t[i]){
			Vg[tw]=4*v[i];
			tw++;
		}
		Vw[tw-1]=4*v[i+1];
	}
	tw--;
	int ta=tw;
	for(int i=n-1;i>=0;i--){
		for(int j=ta;j>=0;j--){
			Vg[j]=min(Vg[j],Vw[ta]+ta-j);
		}
		ta -= 4*t[i];
	}
	for(int i=0;i<=tw;i++)
		Vw[i]=Vg[i];
	ta=0;
	for(int i=0;i<n;i++){
		for(int j=ta;j<=tw;j++){
			Vg[j]=min(Vg[j],Vw[ta]+j-ta);
		}
		ta += 4*t[i];
	}
	ans = 0;
	for(int i=1;i<=tw;i++)
		ans += ((double)Vg[i]+(double)Vg[i-1])/2.0;
	printf("%.10lf\n",ans/16.0);
	return 0;
}