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
//by yjz
#include<bits/stdc++.h>
#define FF first
#define SS second
#define MP make_pair
#define PB push_back
typedef long long ll;
using namespace std;
const int maxn = 200111;
int n;
int a[maxn], b[maxn];
ll ans = 0;
void check(int x)
{
	int s = (b[(x-1+n)%n]+b[(x+1)%n]);
	if (b[x]<a[x])
	{
		puts("-1");
		exit(0);
	}
	if (b[x]-a[x]>=s)
	{
		int tmp = (b[x]-a[x])/s;
		ans += tmp;
		b[x] -= s*tmp;
		check(x);
		check((x-1+n)%n);
		check((x+1)%n);
	}
}
int main()
{
	scanf("%d", &n);
	for (int i=0; i<n; i++) scanf("%d", &a[i]);
	for (int i=0; i<n; i++) scanf("%d", &b[i]);
	for (int i=0; i<n; i++)
	{
		check(i);
	}
	for (int i=0; i<n; i++)
	{
		if (b[i]!=a[i])
		{
			puts("-1");
			return 0;
		}
	}
	cout<<ans<<endl;
	return 0;
}