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;
int n;
int t[105], v[105];
vector <int> lim;
double dp[40005][205];
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; ++ i) scanf("%d", &t[i]);
	for (int i = 0; i < n; ++ i) scanf("%d", &v[i]);
	for (int i = 0; i < n; ++ i) {
		for (int j = 0; j < 2 * t[i]; ++ j) {
			lim.push_back(2 * v[i]);
		}
	}
	for (int i = 0; i <= lim.size(); ++ i) {
		for (int j = 0; j <= 200; ++ j) {
			dp[i][j] = -1e18;
		}
	}
	dp[0][0] = 0;
	for (int i = 0; i < lim.size(); ++ i) {
		for (int j = 0; j <= lim[i]; ++ j) {
			if (j && j - 1 <= lim[i]) dp[i + 1][j - 1] = max(dp[i + 1][j - 1], dp[i][j] + j - 0.5);
			if (j <= lim[i]) dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + j);
			if (j + 1 <= lim[i]) dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + j + 0.5);
		}
	}
	printf("%.9f\n", dp[lim.size()][0] / 4.0);
	return 0;
}