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<iostream>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<math.h>

using namespace std;

#define INF 1000000007
#define LINF (1LL << 62)

typedef long long i64;
typedef pair<i64,i64> P;


int n, t[202], v[202];
vector<int> ma(202020,INF);
double ans = 0;

int main(){
	cin >> n;
	int T = 0;
	for(int i = 0; i < n; i++){
		cin >> t[i];
		T += t[i]*2;
	}
	int now = 0;
	for(int i = 0; i < n; i++){
		cin >> v[i];
		for(int j = 0; j <= t[i]*2; j++){
			ma[now] = min(v[i]*2, ma[now]);
			now++;
		}
		now--;
	}
	
	for(int i = 0; i <= T; i++){
		ma[i] = min(min(T-i,i), ma[i]);
	}
	for(int i = 0; i <= T; i++){
		if(i != 0){
			if(ma[i-1] < ma[i]-1){
				ma[i] = ma[i-1]+1;
			}
			if(ma[T+1-i] < ma[T-i]-1){
				ma[T-i] = ma[T+1-i]+1;
			}
		}
	}
	for(int i = 0; i <= T; i++){
		ans += ma[i];
	}
	printf("%.10f\n", ans/4);
	
	return 0;
}