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 <iostream>
using namespace std;
using ll = long long;

const int N = 2 * (int)1e5;
ll as[N];
ll bs[N];
ll cou = 0;

void check(int i, int n) {
	int pi = (i + (n-1)) % n;
	int ni = (i + 1) % n;
	ll d = bs[i] - as[i];
	if (d > 0) {
		ll add = bs[pi] + bs[ni];
		ll r = d / add;
		cou += r;
		bs[i] -= r * add;

		if (r > 0) {
			check(pi, n);
			check(ni, n);
		}
	}
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	int n;
	cin >> n;

	for (int i = 0; i < n; ++i) cin >> as[i];
	for (int i = 0; i < n; ++i) cin >> bs[i];
	for (int i = 0; i < n; ++i) check(i, n);

	bool fail = false;
	for (int i = 0; i < n; ++i) fail |= (as[i] != bs[i]);

	if (fail) cout << "-1\n";
	else cout << cou << '\n';
}
