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;

typedef long long ll;
const int N=200005;
ll a[N],b[N];
int n;
struct node{
	int id;
	ll val;
	bool operator <(const node &x)const{
		return val<x.val;
	}
};
priority_queue<node> myline;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){scanf("%lld",&a[i]);}
	for(int i=1;i<=n;i++){scanf("%lld",&b[i]);myline.push((node){i,b[i]});}
    bool flag=false;
    ll ans=0;
    while(!myline.empty()){
		node head=myline.top();myline.pop();
		int dd=head.id;
	//	printf("b[%d]=%d,a[%d]=%d\n",dd,b[dd],dd,a[dd]);
		if(a[dd]==b[dd]) continue;
		int l,r;
		if(dd==1) l=n;else l=dd-1;
		if(dd==n) r=1;else r=dd+1; 
		//b[dd]-=b[l];b[dd]-=b[r];
		if(b[dd]-b[l]-b[r]<a[dd]) {puts("-1");flag=true;break;}
		ll tmp=(b[dd]-a[dd])/((b[l]+b[r]));
		b[dd]-=tmp*(b[l]+b[r]);
		ans+=tmp;
		if(b[dd]<a[dd]) {puts("-1");flag=true;break;}
		myline.push((node){dd,b[dd]});
	}
	if(!flag) printf("%lld",ans);
}