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;

#define Debug(x) x
typedef int regular_int;
#define int ll
typedef long long ll;

int x[300000], B[300000];

regular_int main() {
	ios_base::sync_with_stdio(false);
	cout << setprecision(12) << fixed;

	int N;
	cin >> N;

	priority_queue<pair<int, int > > q;
	for(int i = 0; i < N; ++i) {
		cin >> x[i];
	}
	for(int i = 0; i < N; ++i) {
		cin >> B[i];
		if(B[i] < x[i]) {
			cout <<  - 1 << endl;
			return 0;
		}
		if(B[i] > x[i])
			q.push(make_pair(B[i], i));
	}
	int cnt = 0;

	while(!q.empty()) {
		int i = q.top().second;
		q.pop();
		int d = B[(i + 1) % N] + B[(i + N - 1) % N];
		int k = (B[i] - x[i]) / d;
		B[i] -= k * d;
        cnt += k;
		if(!k || B[i] < x[i]) {
			cout <<  - 1 << endl;
			return 0;
		}
		if(B[i] > x[i])
			q.push(make_pair(B[i], i));
	}

	cout << cnt << endl;
	return 0;
}
