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 <bits/stdc++.h>
#define Int int64_t

using namespace std;

int main() {
	int N;
	cin >> N;
	vector<int> A(N), B(N);
	for (int i = 0; i < N; ++i) { cin >> A[i]; }
	for (int i = 0; i < N; ++i) { cin >> B[i]; }

	priority_queue<pair<int, int>> que;
	for (int i = 0; i < N; ++i) {
		if (A[i] < B[i]) {
			que.emplace( B[i], i );
		}
	}

	Int ans = 0;
	while (!que.empty()) {
		auto p = que.top(); que.pop();
		int b = p.first, idx = p.second;
		int l = B[(idx-1+N) % N];
		int r = B[(idx+1) % N];
		
		if ((b - A[idx]) / (l + r) <= 0) { break; }
		ans += (b - A[idx]) / (l + r);
		B[idx] = A[idx] + (b - A[idx]) % (l + r);
		
		if (B[idx] <= A[idx]) { continue; }
		que.emplace( B[idx], idx );
	}

	for (int i = 0; i < N; ++i) {
		if (A[i] == B[i]) { continue; }
		cout << "-1\n";
		return 0;
	}
	cout << ans << "\n";

	return 0;
}
