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;

const int mxN=2e5;
int n, a[mxN], b[mxN];
priority_queue<array<int, 2>> pq;
long long ans;

void fk() {
	cout << -1;
	exit(0);
}

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

	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.size()) {
		int i=pq.top()[1], c=b[(i+1)%n]+b[(i+n-1)%n], t=(b[i]-a[i])/c;
		pq.pop();
		if(a[i]==b[i])
			continue;
		if(t<=0)
			fk();
		b[i]-=t*c;
		pq.push({b[i], i});
		ans+=t;
	}
	cout << ans;
}