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 <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 200005
#define ll long long
int a[N], b[N], n;
priority_queue<pair<int, int> > q;
ll ans;
int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &b[i]);
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]), q.push(make_pair(a[i], i));
	while (!q.empty())
	{
		pair<int, int>t = q.top();
		q.pop();
		if (b[t.second] == t.first) continue;
		if (b[t.second] > t.first)
		{
			puts("-1");
			return 0;
		}
		int dis = t.first - b[t.second];
		int sum = a[t.second - 1] + a[t.second + 1];
		if (t.second == 1)
			sum += a[n];
		if (t.second == n)
			sum += a[1];
		if (sum > dis)
		{
			puts("-1");
			return 0;
		}
		int add = dis / sum;
		ans += add;
		a[t.second] -= add * sum;
		t.first = a[t.second];
		q.push(t);
	}
	printf("%lld\n", ans);
}
