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<cstdio>

int main(){
	int N;
	scanf("%d", &N);
	int t[N], total_t = 0;
	for (int i = 0; i < N; ++i){
		scanf("%d", t + i);
		total_t += t[i];
	}
	int v_max[total_t*2 + 1];
	for (int i = 0; i <= total_t; ++i) v_max[i] = v_max[total_t*2 - i] = i;
	int time_passed = 0;
	for (int i = 0; i < N; ++i){
		int v;
		scanf("%d", &v);
		for (int j = time_passed*2; j < (time_passed + t[i])*2; ++j) if(v_max[j] > v*2) v_max[j] = v*2;
		for (int j = 1; j < time_passed*2; ++j) if(v_max[j] > v*2 + time_passed*2 - j) v_max[j] = v*2 + time_passed*2 - j;
		for (int j = (time_passed + t[i])*2; j < total_t*2; ++j) if(v_max[j] > v*2 + j - (time_passed + t[i])*2) v_max[j] = v*2 + j - (time_passed + t[i])*2;
		time_passed += t[i];
	}
	int answer = 0;
	for (int i = 0; i < total_t*2; ++i) answer += v_max[i] + v_max[i + 1];
	printf("%lf\n", ((double)answer)/8.0);
	return 0;
}