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 ll long long

int main() {
	int N, a, b, c;
	cin >> N;
	vector<ll> A(N), B = A;
	for (int i = 0;i < N;i++) {
		cin >> A[i];
	}
	for (int i = 0;i < N;i++) {
		cin >> B[i];
	}

	queue<int> Q;
	for (int i = 0;i < N;i++) {
		a = i - 1; if (a == -1)a = N - 1;
		b = i;
		c = i + 1; if (c == N)c = 0;

		if (B[b] - A[b] >= B[a] + B[c])Q.push(b);
	}

	ll ans = 0;

	while (!Q.empty()) {
		b = Q.front();
		a = b - 1; if (a == -1)a = N - 1;
		c = b + 1; if (c == N)c = 0;
		Q.pop();

		if (B[b] - A[b] >= B[a] + B[c]) {
			ans += (B[b] - A[b]) / (B[a] + B[c]);
			B[b] = (B[b] - A[b]) % (B[a] + B[c]) + A[b];
			Q.push(a);
			Q.push(c);
		}
	}
	for (int i = 0;i < N;i++) {
		if (A[i] != B[i])ans = -1;
	}
	cout << ans << endl;
}
