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() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
	cin >> n;

	vector<int> a(n);
	for(int i = 0; i < n; ++i) {
		cin >> a[i];
	}

	vector<int> b(n);
	for(int i = 0; i < n; ++i) {
		cin >> b[i];
	}

	int cnt = 0;

	set<pair<int, int> > s = { { 0, -1 } };
	for(int i = 0; i < n; ++i) {
		s.emplace(b[i], i);
	}

	long long ans = 0;

	while(s.size() > 1) {
		auto o = *s.rbegin();
		s.erase(o);
		int p = o.second;
		int nxt = s.rbegin() -> first;
		int m = b[(p - 1 + n) % n] + b[(p + 1) % n];
		int delta = (o.first - a[p]) / m;
		if(o.first == a[p]) {
			continue;
		}
		if(delta == 0 || o.first < a[p]) {
			cout << -1 << '\n';
			exit(0);
		}
		ans += delta;
		o.first -= m * delta;
		b[p] = o.first;
		s.insert(o);
	}

	cout << ans << '\n';

	return 0;
}