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;
const int maxn = 111;
int t[maxn], v[maxn];
int mx[maxn * maxn * 20];
int main(){
	int n; cin >> n;
	for(int i = 1; i <= n; i++) cin >> t[i];
	for(int i = 1; i <= n; i++) cin >> v[i];
	n += 2;
	for(int i = 0; i < n; i++) t[i] *= 2, v[i] *= 2;
	int sum = 0;
	for(int i = 0; i < n; i++) sum += t[i];
	for(int i = 0; i < sum; i++) mx[i] = 1<<30;
	int now = 0;
	for(int i = 0; i < n; i++){
		for(int j = 0; j <= t[i]; j++){
			mx[now + j] = min(mx[now + j], v[i]);
		}
		now += t[i];
	}
	mx[0] = 0;
	mx[sum] = 0;
	for(int i = 0; i < sum; i++) if(i) mx[i] = min(mx[i], mx[i - 1] + 1);
	for(int i = sum - 1; i >= 0; i--) mx[i] = min(mx[i], mx[i + 1] + 1);
	double ans = 0;
	for(int i = 0; i < sum; i++) ans += mx[i] + mx[i + 1];
	cout << fixed << setprecision(12) << ans / 8.0 << endl;
	return 0;
}