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;
using ll = long long;

//https://atcoder.jp/contests/agc037/tasks/agc037_c

const int nm = 2e5;
int n, a[nm], b[nm];
ll ans = 0;

priority_queue<pair<int, int>> q;

int main() {
	cin >> n;
	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) {
		if (b[i] < a[i]) {
			cout << -1 << endl;
			return 0;
		}
		if (b[i] != a[i])
			q.emplace(b[i], i);
	}
	while (!q.empty()) {
		int i = q.top().second;
		q.pop();

		int t = (b[i] - a[i]) / (b[(i + n - 1) % n] + b[(i + 1) % n]);
		if (!t) {
			cout << -1 << endl;
			return 0;
		}
		ans += t;
		b[i] -= (b[(i + n - 1) % n] + b[(i + 1) % n]) * t;

		if (b[i] != a[i])
			q.emplace(b[i], i);
	}
	cout << ans << endl;
}
