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 <cstdio>
#include <queue>
const int N = 3e5;
int a[N], b[N];

struct data 
{
	int id;
	bool operator < (const data &other) const
	{
		return b[id] < b[other.id];
	}
};

int main()
{
	int n; scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &a[i]);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &b[i]);
	std::priority_queue<data> q;
	for (int i = 1; i <= n; ++i)
		if (b[i] > a[i])
			q.push({i});
	long long res = 0;
	while (!q.empty())
	{
		int x = q.top().id;
		q.pop();
		int prv = x > 1 ? x - 1 : n;
		int nxt = x < n ? x + 1 : 1;
		int sum = b[prv] + b[nxt];
		int cnt = (b[x] - a[x]) / sum;
		if (cnt == 0)
			break;
		res += cnt;
		b[x] -= sum * cnt;
		if (b[x] > a[x])
			q.push({x});
	}
	for (int i = 1; i <= n; ++i)
		if (a[i] != b[i])
		{
			printf("-1");
			return 0;
		}
	printf("%lld\n", res);
}