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>
#define int long long
#define fi first
#define se second
using namespace std;
const int N = 2e5 + 3;
int a[N], b[N];
priority_queue <pair <int,int> > pq;
signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}

	for(int i = 1; i <= n; i++)
	{
		cin >> b[i];
		if(b[i] < a[i])
			return cout << -1, 0;
		else if(b[i] > a[i])
			pq.push({b[i], i});
		
	}
	b[0] = b[n], b[n + 1] = b[1];
	int ans = 0;
	while(pq.size() > 0)
	{
		int i = pq.top().se;
		pq.pop();
		if(b[i] < b[i + 1] + b[i - 1])
			return cout << -1, 0;

		int tmp = (b[i + 1] + b[i - 1]);
		int step = (b[i] - a[i])/tmp;
		if(step == 0)
			return cout << -1, 0;
		ans += step;
		b[i] -= step * tmp;
		b[0] = b[n];
		b[n + 1] = b[1];

		if(b[i] > a[i])
			pq.push({b[i], i});
	}
	cout << ans;
}