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 int long long
#define fi first
#define se second
const int N=2e5+5;
priority_queue< pair<int,int> > q;
int n,c[N],a[N];
int le(int u){
	return u==1?n:u-1;
}
int re(int u){
	return u==n?1:u+1;
}
signed main(){
	///freopen("in01.txt","r",stdin);
	cin>>n;
	for(int i=1;i<=n;++i)
		cin>>a[i];
	for(int i=1;i<=n;++i)
		cin>>c[i];
	for(int i=1;i<=n;++i)
		if(c[i]!=a[i]) q.push(make_pair(c[i],i));
	int ans=0;
	while(!q.empty()){
		int u=q.top().se,tmp=0; q.pop();
		tmp=c[le(u)]+c[re(u)];
		if(q.empty()){
			if((c[u]-a[u])%tmp!=0) return puts("-1"),0;
			ans+=(c[u]-a[u])/tmp; c[u]=a[u]; break;
		}
		int t=(c[u]-a[u])/tmp;t=max(t,1ll);
		c[u]-=t*tmp; ans+=t;
		if(c[u]<a[u]) return puts("-1"),0;
		if(c[u]>a[u]) q.push(make_pair(c[u],u));
	}
	cout<<ans;
	return 0;
}
