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 main() {
	int N, L, R;
	double count = 0, ans = 0;
	pair<double, double> P[100];
	map<double, double> mpL, mpR, mp;
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> P[i].first;
		count += P[i].first;
		if (i != 0) P[i].first += P[i - 1].first;
	}
	for (int i = 0; i < N; i++) {
		cin >> P[i].second;
	}
	L = 0, R = N - 1;
	mpL[0] = 0, mpR[count] = 0;
	for (double i = 0.5; i <= count; i += 0.5) {
		double maxS = P[L].second;
		mpL[i] = mpL[i - 0.5] + 0.5;
		mpL[i] = min(mpL[i], maxS);
		if (i == P[L].first) L++;
	}
	for (double i = count - 0.5; i >= 0; i -= 0.5) {
		double maxS = P[R].second;
		mpR[i] = mpR[i + 0.5] + 0.5;
		mpR[i] = min(mpR[i], maxS);
		if (R != 0 && i == P[R - 1].first) R--;
	}
	for (double i = 0; i <= count; i += 0.5) {
		mp[i] = min(mpL[i], mpR[i]);
	}
	for (double i = 0; i < count; i += 0.5) {
		ans += (mp[i] + mp[i + 0.5]) / 4;
	}
	printf("%.8f", ans);
}