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;
const int N = 200005;
int a[N],b[N],n;
priority_queue<pair<int ,int > >q;
int fro(int x){return x==1?n:x-1;}
int bac(int x){return x==n?1:x+1;}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)scanf("%d",&b[i]),q.push(make_pair(b[i],i));
	long long tmp=0;
	while(!q.empty())
	{
		int i=q.top().second;q.pop();if(b[i]==a[i])continue;
		if(b[i]-a[i]<b[fro(i)]+b[bac(i)])return puts("-1"),0;
		int sn = b[fro(i)]+b[bac(i)],lim;
		lim = b[i] - a[i],tmp+=lim/sn,lim%=sn,b[i]=a[i]+lim;
		if(b[i]<a[i])return puts("-1"),0;
		if(b[i]!=a[i])q.push(make_pair(b[i],i));
	}
	printf("%lld\n",tmp);
}
