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>



double dy[110][210][110];
bool chk[110][210][110];
int T[110];
int V[110];
int n;

double go(int idx, int t, int v) {
	if (idx == n && v == 0) {
		return 0;
	}
	if (idx == n) {
		return -1e9;
	}
	if (v > V[idx])return -1e9;
	auto& ret = dy[idx][t][v];
	if (chk[idx][t][v])return ret;
	chk[idx][t][v] = true;
	ret = -1e9;
	if (t == 0) {
		ret = go(idx + 1, T[idx+1], v);
		return ret;
	}
	if (v < V[idx]) {
		ret = std::max(ret, go(idx, t - 1, v + 1) + v + 0.5);
		ret = std::max(ret, go(idx, t - 1, v) + v + 0.25);
	}
	ret = std::max(ret, go(idx, t - 1, v) + v);
	if (v > 0) {
		ret = std::max(ret, go(idx, t - 1, v - 1) + v - 0.5);
	}
	return ret;
}
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]);
	}
	printf("%.10lf", go(0, T[0], 0));
}