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 BUF (105)
#define INF (1000000006)
#define ll long long 

int n;
int t[BUF];
int v[BUF];
int g[100*200*2+5];

int f(int x, int y){
	if(y==0) return x;
	if(y==n+1) return -x+2*t[n];

	if(x<2*t[y-1]){
		return -x+2*t[y-1]+2*v[y];
	}else if(x>2*t[y]){
		return x-2*t[y]+2*v[y];
	}else {
		return 2*v[y];
	}
}




int main(void){
	cin >> n;
	int tm=0;

	for(int i=1;i<=n;i++){
		int tmp;
		cin >> tmp;
		tm+=tmp;
		t[i]=tm;
	}

	for(int i=1;i<=n;i++){
		cin >> v[i];
	}


	for(int i=0;i<=2*t[n];i++){
		int m = INF;
		for(int k=0;k<=n+1;k++){
			m=min(m,f(i,k));
		}
		g[i]=m;
		//cout << m << " ";
	}

	int ans = 0;
	for(int i=0;i<2*t[n];i++){
		ans += g[i]+g[i+1];
	}

	cout << setprecision(20) <<  (double)ans/(double)8 << endl;
	return 0;
}