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 rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef pair<int,int>P;
typedef long long ll;

int a[300000],b[300000];
int main(){
	int n;cin>>n;
	rep(i,n){
		scanf("%d",&a[i]);
	}
	priority_queue<P>que;
	rep(i,n){
		scanf("%d",&b[i]);
		if(b[i]!=a[i]){
			if(b[i]<a[i]){
				puts("-1");return 0;
			}
			que.push(P(b[i],i));
		}
	}
	ll cnt=0;
	while(!que.empty()){
		P p=que.top();que.pop();
		int l=(p.second+n-1)%n,r=(p.second+1)%n;
		int x=p.second;
		int need=b[x]-a[x];
		if(need<(b[l]+b[r])){
			puts("-1");return 0;
		}
		else{
			int d=need/(b[l]+b[r]);
			cnt+=d;
			b[x]-=d*(b[l]+b[r]);
			if(b[x]!=a[x]){
				que.push(P(b[x],x));
			}
		}
	}
	cout<<cnt<<endl;
}