There are N positive integers arranged in a circle.

Now, the i-th number is A_i. Takahashi wants the i-th number to be B_i. For this objective, he will repeatedly perform the following operation:

* Choose an integer i such that 1 \leq i \leq N.
* Let a, b, c be the (i-1)-th, i-th, and (i+1)-th numbers, respectively. Replace the i-th number with a+b+c.



Here the 0-th number is the N-th number, and the (N+1)-th number is the 1-st number.

Determine if Takahashi can achieve his objective. If the answer is yes, find the minimum number of operations required.

Constraints

* 3 \leq N \leq 2 \times 10^5
* 1 \leq A_i, B_i \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N
B_1 B_2 ... B_N


Output

Print the minimum number of operations required, or `-1` if the objective cannot be achieved.

Examples

Input

3
1 1 1
13 5 7


Output

4


Input

4
1 2 3 4
2 3 4 5


Output

-1


Input

5
5 6 5 2 1
9817 1108 6890 4343 8704


Output

25
#include <iostream>
#include <vector>
#include <tuple>
#include <queue>

void failed () {
	std::cout << -1 << '\n';
	exit(0);
}

int main() {
	int n;
	std::cin >> n;
	std::vector<int> a(n), b(n);
	for (int &e : a) std::cin >> e;
	for (int &e : b) std::cin >> e;
	
	long long ans = 0;
	std::priority_queue<std::pair<int, int>> que;
	for (int i = 0; i < n; i++) {
		if (a[i] != b[i]) que.emplace(b[i], i);
		if (a[i] > b[i]) failed();
	}
	
	while (not que.empty()) {
		auto [value, i] = que.top(); que.pop();
		const int l = (i == n - 1 ? 0 : i + 1);
		const int r = (i == 0 ? n - 1 : i - 1);
		const int s = b[l] + b[r];
		if (b[i] < s) failed();
		
		const int count = (b[i] - std::max(s, a[i]) + s - 1) / s;
		if ((b[i] -= (count * s)) < a[i]) failed();
		if (b[i] != a[i]) que.emplace(b[i], i);
		ans += count;
	}
	
	std::cout << ans << '\n';
	
	return 0;
}