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>
#define int long long
#define mid ((l+r)/2)
using namespace std;
int n,ans,a[222222],b[222222],t[888888];
void update(int l,int r,int p,int c,int v){
	if(l==r){t[p]=v;return;}
	mid>=c?update(l,mid,p*2,c,v):update(mid+1,r,p*2+1,c,v);
	t[p]=max(t[p*2],t[p*2+1]);
}
int query(int l,int r,int p){
	if(l==r)return l;
	return t[p]==t[p*2]?query(l,mid,p*2):query(mid+1,r,p*2+1);
}
signed 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]),update(1,n,1,i,b[i]);
	b[0]=b[n],b[n+1]=b[1];
	for(int tmp=n;tmp;){
		int p=query(1,n,1),d=b[p]-a[p],k=b[p+1]+b[p-1];
		if(!d){tmp--,update(1,n,1,p,0);continue;}
		if(d<k){printf("-1");return 0;}
		ans+=d/k,b[p]-=d/k*k,b[0]=b[n],b[n+1]=b[1],update(1,n,1,p,b[p]==a[p]?0:b[p]);
		if(b[p]==a[p])tmp--;
	}
	printf("%lld",ans);
}