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 <utility>
#include <queue>

using namespace std;

int main() {
	int n; cin >> n;
	vector<int> a(n), b(n);
	priority_queue<pair<int, int> > pq;
	for (int i = 0; i < n; ++i) cin >> a[i];
	for (int i = 0; i < n; ++i) cin >> b[i];
	for (int i = 0; i < n; ++i) pq.push(make_pair(b[i], i));
	long long ans = 0;
	while (!pq.empty()) {
		int now = pq.top().second; pq.pop();
		int dif = b[now] - a[now];
		int tmp = b[(now - 1 + n) % n] + b[(now + 1) % n];
		if (dif != 0 && dif < tmp) {
			ans = -1;
			break;
		}
		ans += dif / tmp;
		b[now] -= dif / tmp * tmp;
		if (b[now] != a[now]) pq.push(make_pair(b[now], now));
	}
	cout << ans << endl;

	return 0;
}