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 pair<int, int> pii;

int a[212123];
int b[212123];
int n;

int main() {
	priority_queue<pii> pq;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	for (int i = 0; i < n; i++) {
		cin >> b[i];
		pq.emplace(b[i], i);
	}
	long long ans = 0;
	while (pq.size()) {
		int x, idx;
		tie(x, idx) = pq.top();
		pq.pop();
		if (b[idx] == a[idx]) continue;
		if (b[idx] < a[idx]) goto bad;

		int l = (idx + n - 1) % n;
		int r = (idx + 1) % n;
		int s = b[l] + b[r];
		if (b[idx] <= s) goto bad;
		// check if a[idx] can be gotten to
		if ((b[idx] - a[idx]) % s == 0) {
			ans += (b[idx] - a[idx]) / s;
			b[idx] = a[idx];
			continue;
		}

		ans += b[idx] / s;
		b[idx] %= s;

		if (b[idx] < a[idx]) goto bad;
		if (b[idx] > a[idx]) pq.emplace(b[idx], idx);
	}
	cout << ans << '\n';
	return 0;

bad:
	cout << "-1\n";
	return 0;
}

