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() {
	long long N, ANS = 0;
	static long long A[200000], B[200000];
	priority_queue<pair<long long, int> > PQ;
	cin >> N;
	for (int i = 0; i < N; i++) cin >> A[i];
	for (int i = 0; i < N; i++) cin >> B[i], PQ.push({ B[i], i });
	while (!PQ.empty()) {
		pair<long long, int> P = PQ.top();
		PQ.pop();
		long long X = P.first, Y = P.second;
		long long L = B[(Y + N - 1) % N], R = B[(Y + 1) % N];
		if (X  - A[Y] < L + R) continue;
		long long Diff = X - A[Y];
		long long Z = Diff / (L + R);
		B[Y] -= Z * (L + R);
		ANS += Z;
		PQ.push({ B[Y], Y });
	}
	for (int i = 0; i < N; i++) {
		if (A[i] != B[i]) {
			cout << -1 << endl;
			return 0;
		}
	}
	cout << ANS << endl;
}