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 int long long
using namespace std;
const int N=2e5+10;
int n,a[N],b[N],ans;
priority_queue<pair<int,int> >q;
signed main() {
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	for(int i=1;i<=n;i++){
		scanf("%lld",&b[i]);q.push(make_pair(b[i],i));	
	}
	while(q.size()){
		pair<int,int> x=q.top();q.pop();
		if(a[x.second]==x.first)continue;
		if(a[x.second]>x.first){
			return puts("-1")*0;
		}
		int dis=x.first-a[x.second];
		int sum=b[x.second-1]+b[x.second+1];
		if(x.second==1)sum+=b[n];
		if(x.second==n)sum+=b[1];
		if(sum>dis){
			return puts("-1")*0;
		}
		int Dis=dis/sum;ans+=Dis;
		b[x.second]-=Dis*sum;
		q.push(make_pair(b[x.second],x.second));
	}
	cout<<ans<<endl;
	return 0;
}
