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 main() {
	int n; cin >> n;
	int64_t a[n]; for(int i = 0; i < n; ++i) cin >> a[i];
	int64_t b[n]; for(int i = 0; i < n; ++i) cin >> b[i];
	priority_queue<pair<int, int>, vector<pair<int, int>>> pq;
	for(int i = 0; i < n; ++i) {
		pq.emplace(b[i], i);
	}
	int64_t ans = 0;
	while(not pq.empty()) {
		int _, i;
		tie(_, i) = pq.top();
		pq.pop();
		if(b[i] == a[i]) continue;
		int64_t l = b[(i - 1 + n) % n];
		int64_t r = b[(i + 1) % n];
		if(b[i] <= l + r) {
			cout << -1 << '\n';
			return 0;
		}
		if((b[i] - a[i]) % (l + r) == 0) {
			ans += (b[i] - a[i]) / (l + r);
			b[i] = a[i];
			continue;
		}
		int k = (b[i] + (l + r) - 1) / (l + r) - 1;
		ans += k;
		b[i] = b[i] - k * (l + r);
		pq.emplace(b[i], i);
	}
	cout << ans << '\n';
	return 0;
}