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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int a[202020], b[202020];
typedef long long ll;
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num; i++)scanf("%d", &a[i]);
	for (int i = 0; i < num; i++)scanf("%d", &b[i]);
	vector<int>v;
	for (int i = 0; i < num; i++)if (b[(i + num - 1) % num] + b[(i + 1) % num] <= b[i] - a[i])v.push_back(i);
	ll ans = 0;
	for (;;)
	{
		if (v.empty())break;
		int z = v[v.size() - 1];
		v.pop_back();
		int s = (z + num - 1) % num, t = (z + 1) % num;
		if (b[s] + b[t] <= b[z] - a[z])
		{
			int x = (b[z] - a[z]) / (b[s] + b[t]);
			ans += x;
			b[z] -= (b[s] + b[t])*x;
			v.push_back(s), v.push_back(t);
		}
	}
	for (int i = 0; i < num; i++)
	{
		if (a[i] != b[i])
		{
			printf("-1\n");
			return 0;
		}
	}
	printf("%lld\n", ans);
}