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 ll long long
using namespace std;
const ll N=2e5+10;
ll n,ans;
ll a[N],b[N];
struct node{
	ll rt,val;
	bool operator <(const node & op) const{ return val<op.val; }	
};
priority_queue <node> pq;
int main(){
	scanf("%lld",&n);
	for(ll i=1;i<=n;i++) scanf("%lld",&a[i]);
	for(ll i=1;i<=n;i++){
		scanf("%lld",&b[i]);
		if(a[i]!=b[i]) pq.push((node){i,b[i]});
	}

	while(!pq.empty()){
		ll pt=pq.top().rt,pre=(pt-2+n)%n+1,suf=pt%n+1;
		pq.pop();
		ll opt=(b[pt]-a[pt])/(b[pre]+b[suf]);
		if(opt==0){puts("-1");exit(0);}
		b[pt]-=opt*(b[pre]+b[suf]);
		ans+=opt;
		if(a[pt]!=b[pt]) pq.push((node){pt,b[pt]});
	}
	printf("%lld",ans);
}