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;
#define N 200005
#define LL long long
#define pii pair<int,int>
#define mp make_pair

int n,ci;
LL ans;
int a[N],b[N];
priority_queue<pii> q;

int main()
{
	ios_base::sync_with_stdio(0);
	
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	for(int i=0;i<n;i++)
		cin>>b[i];
	for(int i=0;i<n;i++)
		q.push(mp(b[i],i));
	while(!q.empty())
	{
		ci++;
		if(ci>=1e7) return cout<<"-1\n",0;
		int x=q.top().second;
		q.pop();
		if(b[x]==a[x]) continue;
		int pa,pb;
		if(x) pa=b[x-1];
		else pa=b[n-1];
		if(x!=n-1) pb=b[x+1];
		else pb=b[0];
		if(max(max(pa,pb),a[x])+pa+pb<b[x])
		{
			ans+=(b[x]-max(max(pa,pb),a[x]))/(pa+pb);
			b[x]-=(b[x]-max(max(pa,pb),a[x]))/(pa+pb)*(pa+pb);
		}
		else b[x]-=pa+pb,ans++;
		if(b[x]<a[x]) return cout<<"-1\n",0;
		q.push(mp(b[x],x));
	}
	cout<<ans<<'\n';
	
	return 0;
}