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;

int n, a[200005], b[200005];
bool no[200005];
long long ans;
priority_queue<pair<int, int>> q;

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%d", a + i);
	for (int i = 0; i < n; i++) {
		scanf("%d", b + i);
		q.emplace(b[i], i);
	}
	while (!q.empty()) {
		int ta = q.top().first;
		int tb = q.top().second;
		q.pop();
		int tc = (tb + 1) % n;
		int td = (tb + n - 1) % n;
		int sum = b[tc] + b[td];
		if (ta <= a[tb])
			continue;
		int te = (ta - a[tb]) / sum;
		if (te) {
			ans += te;
			b[tb] -= te * sum;
			q.emplace(b[tb], tb);
			if (no[tc]) {
				q.emplace(b[tc], tc);
				no[tc] = 0;
			}
			if (no[td]) {
				q.emplace(b[td], tc);
				no[td] = 0;
			}
		} else
			no[tb] = 1;
	}
	for (int i = 0; i < n; i++)
		if (a[i] != b[i]) {
			puts("-1");
			return 0;
		}
	printf("%lld\n", ans);
}
