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;
typedef long long ll;
const int maxn=2e5+5;
 
priority_queue<pair<int,int> >pq;
 
int n;
int a[maxn],b[maxn];
ll ans=0;
 
inline int getpos(const int &x){if(x==n+1)return 1;if(x==0)return n;return x;}
void No(){cout<<"-1\n";exit(0);}
 
int main()
{
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	
//	freopen("data.in","r",stdin);
	
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=1;i<=n;i++)cin>>b[i];
	for(int i=1;i<=n;i++)
	{
		pq.push(make_pair(b[i],i));
	}
	
	while(!pq.empty())
	{
		int p=pq.top().second;pq.pop();
		int minu=b[getpos(p-1)]+b[getpos(p+1)];
		int k=(b[p]-a[p])/minu;
		if((b[p]-a[p])%minu==0)
		{
			ans+=k;
			b[p]=a[p];
			continue;
		}
		if(minu>=b[p])No();
		if(k==0)k++;
		ans+=k;
		b[p]%=minu;pq.push(make_pair(b[p],p));
		if(b[p]<a[p])No();
	}
	cout<<ans<<'\n';
//	cerr<<"loopcount:"<<lc<<'\n';
	return 0;
}