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 fio() ios_base::sync_with_stdio(false); cin.tie(NULL)
#define int long long
#define MOD 998244353
using namespace std;
int32_t main()
{
	fio();
	int n;
	cin>>n;
	vector<int>a(n),b(n);
	priority_queue<pair<int,int> >pq;

	for(int i =0;i<n;i++)cin>>a[i];
	for(int i =0;i<n;i++)
	{
		cin>>b[i];
		if(b[i]!=a[i])
		pq.push({b[i],i});
	}
	int ans = 0;int f = 1;
	while(!pq.empty())
	{
		int val = pq.top().first;
		int idx = pq.top().second;
		pq.pop();
		int sm = b[(idx+1)%n] + b[(idx-1+n)%n];
		ans+=(val-a[idx])/sm;
		int lol = (val-a[idx])/sm;
		if((val-a[idx])/sm <= 0)
		{
			f = 0;
			break;
		}
		b[idx] -=lol*sm;
		if(b[idx]>a[idx])
			pq.push({b[idx],idx});
	}
	if(f==1)
		cout<<ans;
	else cout<<-1;
}