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>
#include <algorithm>
#include <cstring>
#define inf 0x7fffffff
using namespace std;

int n, t[10005], v[10005], sumt;
double ans, limt[40005];
int main() {
	for(int i = 0; i < 40005; i++) limt[i] = inf;
	ans = 0.0;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
	scanf("%d", &t[i]), sumt += t[i];
	sumt *= 2;
	for(int i = 1; i <= n; i++)
	scanf("%d", &v[i]);
	double vs = 0.0;
	int ts = 0;
	for(int i = 1; i <= n; i++) {
		for(int k = ts; k <= ts + 2 * t[i]; k++)
		limt[k] = min(limt[k], 1.0 * v[i]);
		double p = 0.5;
		for(int k = ts - 1; k >= 0; k--, p += 0.5)
		limt[k] = min(limt[k], 1.0 * v[i] + p);
		ts += 2 * t[i];
	}
	double p = 0;
	for(int i = sumt; i >= 0; i--, p += 0.5)
	limt[i] = min(limt[i], p);
	for(int i = 1; i <= sumt; i++) {
		double vn = min(vs + 0.5, limt[i]);
		ans += (vs + vn) * 0.25;
		vs = vn;
	}
	printf("%lf\n", ans);
}