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 rep2(x,fr,to) for(int (x)=(fr);(x)<(to);(x)++)
#define rep(x,to) for(int (x)=0;(x)<(to);(x)++)
#define repr(x,fr,to) for(int (x)=(fr);(x)>=(to);(x)--)
#define all(c) (c).begin(),(c).end()
#define sz(v) (int)(v).size()
 
typedef int64_t ll; typedef vector<int> VI; typedef pair<int,int> pii;
const ll mod = 1e9+7;


int main()
{
 
	//cin.tie(0); ios_base::sync_with_stdio(false);
	int n;
	cin >>n;
	VI t(n), v(n);
	rep(i,n) cin >> t[i];
	rep(i,n) cin >> v[i];
	
	vector<double> dp(105, -1);
	dp[0] =0;
	rep(i,n) rep(j, t[i]){
		vector<double> np(105, -1);
		rep(k, v[i]+1) if(dp[k] >=0){
			np[k] = max(np[k], dp[k] +k);
			if(k < v[i]){
				np[k]   = max(np[k], dp[k]+k+0.25);
				np[k+1] = max(np[k+1], dp[k]+k+0.5);
			}
			if(k>0) np[k-1] = max(np[k-1], dp[k]+k-0.5);
		
		}
			swap(dp, np);
	}
			printf("%.10f\n", dp[0]);

	return 0;
}