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>
using namespace std;
typedef long long ll;

int A[200005], B[200005];

inline int mod(int i, int m) { return i >= m ? i - m : i; }

int main() {
	int N;
	cin >> 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>> pq;
	for (int i = 0; i < N; ++i) {
		if (A[i] < B[i]) {
			pq.emplace(B[i], i);
		}
		else if (A[i] > B[i]) {
			cout << -1 << '\n';
			return 0;
		}
	}

	ll ans = 0;
	while (!pq.empty()) {
		pair<int, int> p = pq.top();
		pq.pop();
		int i = p.second, d = B[mod(i - 1 + N, N)] + B[mod(i + 1, N)];
		if (B[i] - A[i] < d) {
			cout << -1 << '\n';
			return 0;
		}
		ans += (B[i] - A[i]) / d;
		B[i] -= d * ((B[i] - A[i]) / d);
		if (A[i] < B[i]) {
			pq.emplace(B[i], i);
		}
	}
	cout << ans << '\n';
	return 0;
}
