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 maxn=200100;
int n,a[maxn],b[maxn];
set<pair<int,int> > S;
inline int prev(int x){return x==1?n:x-1;}
inline int nxtv(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);
	for(int i=1;i<=n;i++) S.insert(make_pair(b[i],i));
	long long ans=0;
	while(S.size())
	{
		pair<int,int> ttt=*--S.end();S.erase(--S.end());
		int id=ttt.second;if(a[id]==ttt.first) continue;
		int cnt=(b[id]-a[id])/(b[prev(id)]+b[nxtv(id)]);
		if(cnt==0) return puts("-1"),0;
		b[id]-=cnt*(b[prev(id)]+b[nxtv(id)]);
		S.insert(make_pair(b[id],id));ans+=cnt;
	}
	cout<<ans<<endl;
	return 0;
}