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<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int n;
#define Maxn 200010
int a[Maxn],b[Maxn];
struct Data{
	int id,val;
	bool operator <(const Data &z)const{return val<z.val;}
};
priority_queue<Data> Q;
int main(){
	scanf("%d",&n);
	for(register int i=0;i<n;++i)scanf("%d",&b[i]);
	for(register int i=0;i<n;++i){
	    scanf("%d",&a[i]);
	    if(a[i]<b[i]){
	    	puts("-1");
	    	return 0;
		} 
	}
	for(register int i=0;i<n;++i)
	if(a[i]!=b[i])Q.push((Data){i,a[i]});
	long long ans=0;
	while(!Q.empty()){
		int x=Q.top().id;
		Q.pop();
		int y1=(x-1+n)%n,y2=(x+1)%n;
	    int tmp=(a[x]-b[x])/(a[y1]+a[y2]);
	    ans+=tmp;
	    if(tmp==0&&a[x]!=b[x]){puts("-1");return 0;}
	    a[x]-=tmp*(a[y1]+a[y2]);
	    if(a[x]!=b[x])Q.push((Data){x,a[x]});
	}
	printf("%lld\n",ans);
	return 0;
}