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;
long long n,a[200010],b[200010];
long long ans=0;
bool w=false;
struct node{
	long long pos,val;
	bool operator <(const node &b)const{
		return val<b.val;
	}
};
priority_queue<node>q;
int main(){
	scanf("%lld",&n);
	for(long long i=1;i<=n;i++) scanf("%lld",&a[i]);
	for(long long i=1;i<=n;i++){
		scanf("%lld",&b[i]);
		node tmp={i,b[i]};
		q.push(tmp);
	}
	while(q.size()){
		node now=q.top();q.pop();
		long long l=(now.pos-2+n)%n+1,r=now.pos%n+1;
		if (b[now.pos]==a[now.pos]) continue;
		if (b[now.pos]<a[now.pos])
		{
			puts("-1");
			return 0;
		}
		if(b[now.pos]-b[l]-b[r]<a[now.pos]){
			puts("-1");
			return 0;
		}
		long long d=b[now.pos]-a[now.pos];
		ans+=d/(b[l]+b[r]);
		b[now.pos]=d%(b[l]+b[r])+a[now.pos];
		if(b[now.pos]==a[now.pos]) continue;
		node tmp={now.pos,b[now.pos]};
		q.push(tmp);
	}
	printf("%lld\n",ans); 
}
