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>
using namespace std;

int main(int argc, char const *argv[])
{
	int n;
	cin>>n;
	long long a[n],b[n];
	multiset<pair<long long,int>> memo;
	for (int i = 0; i < n; ++i)
	{
		cin>>a[i];
	}
	for (int i = 0; i < n; ++i)
	{
		cin>>b[i];
		if (a[i]!=b[i])
		{
			memo.insert(make_pair(b[i],i));
		}
	}
	long long ans=0LL;
	while (memo.size()>0)
	{
		auto i=memo.end();
		--i;
		auto p=*i;
		int l=(p.second-1+n)%n,r=(p.second+1)%n;
		memo.erase(i);
		b[p.second]-=a[p.second];
		if (b[p.second]/(b[l]+b[r])==0)
		{
			cout<<"-1\n";
			return 0;
		}
		ans+=b[p.second]/(b[l]+b[r]);
		b[p.second]%=(b[l]+b[r]);
		b[p.second]+=a[p.second];
		if (a[p.second]!=b[p.second])
		{
			memo.insert(make_pair(b[p.second],p.second));
		}
	}
	cout<<ans<<endl;
	return 0;
}
