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 unsigned long long ull;
typedef long long ll;

#define MAX 100

int t[MAX+2], v[MAX+2], s[MAX+2];

double f(int i, double t){
	if(i>0 && t<=s[i-1]) return v[i]+(s[i-1]-t);
	else if(t<=s[i]) return v[i];
	else return v[i]+(t-s[i]);
}

int main(){
	int n;
	cin >> n;
	t[0]=0, v[0]=0;
	t[n+1]=0, v[n+1]=0;
	for(int i=1; i<=n; i++) cin >> t[i];
	for(int i=1; i<=n; i++) cin >> v[i];

	s[0]=t[0];
	for(int i=1; i<=n+1; i++) s[i]=s[i-1]+t[i];

	double ans=0.0;
	for(int i=0; i<2*s[n+1]; i++){
		double f1=100.0, f2=100.0;
		for(int j=0; j<=n+1; j++) f1=min(f1, f(j, i/2.0));
		for(int j=0; j<=n+1; j++) f2=min(f2, f(j, (i+1)/2.0));
		ans+=(f1+f2)/4.0;
	}
	printf("%.9f\n", ans);
	return 0;
}