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
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define INF (1<<30)
#define INFLL (1ll<<60)
typedef pair<int, int> P;
#define MOD (1000000007ll)
#define l_ength size
#define EPS (1e-10)

ll g[44444];

int main(void){
	ll i,j,n,t[123],v[123],ans=0ll;
	cin >> n;
	t[0] = 0ll;
	for(i=1ll; i<=n; ++i){
		cin >> t[i];
		t[i] *= 2;
		t[i] += t[i-1];
	}
	for(i=0ll; i<n; ++i){
		cin >> v[i];
		v[i] *= 2;
	}
	for(i=0ll; i<=t[n]; ++i){
		g[i] = min(i,t[n]-i);
		for(j=0ll; j<n; ++j){
			if(i <= t[j]){
				g[i] = min(g[i],v[j]+t[j]-i);
			}else if(t[j+1] <= i){
				g[i] = min(g[i],v[j]+i-t[j+1]);
			}else{
				g[i] = min(g[i],v[j]);
			}
		}
	}
	for(i=0ll; i<t[n]; ++i){
		ans += g[i]+g[i+1];
	}
	printf("%.5lf\n",((double)(ans)/8));
	return 0;
}
