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;
#define rep(i,n) for(int i = 0;i < n;i++)
#define rrep(i,n) for(int i = 1;i <= n;i++)

double t[204], v[104], T = 0;
double ok(double t){
	return min(t, T-t);
}

double check(double x, int k){
	if(t[k-1] <= x && x <= t[k])return v[k];
	if(x < t[k])return (t[k-1]-x)+v[k];
	else return (x-t[k])+v[k];
	return 0;
}

int main(){
	int n;
	scanf("%d", &n);
	rrep(i,n)scanf("%lf", &t[i]);
	rrep(i,n)scanf("%lf", &v[i]);
	rrep(i,n)T += t[i];
	rrep(i,n)t[i+1] += t[i];
	
	double vec = 0, ans = 0;
	for(double t = 0;t <= T;t += 0.5){
		double res = (double)1e9;
		rrep(i,n)res = min(res, check(t, i));
		res = min(res, ok(t));
		ans += 0.5*vec+(res-vec)*0.5/2.0;
		vec = res;
	}
	
	printf("%lf\n", ans);
	return 0;
}