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 <algorithm>
#include <queue>

using namespace std;

const int MN = 200005;

using ll = long long;

ll a[MN];
ll b[MN];
int n;
queue<int> q;

void trypush(int i){
	if((b[i]>b[(i-1+n)%n]+b[(i+1)%n])&&b[i]>a[i])q.push(i);
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin>>n;
	for(int i=0; i<n; ++i)cin>>a[i];
	for(int i=0; i<n; ++i)cin>>b[i];
	for(int i=0; i<n; ++i){
		trypush(i);
	}
	ll res=0;
	while(!q.empty()){
		int i=q.front();
		q.pop();
		ll sum = b[(i+1)%n]+b[(i+n-1)%n];
		ll hm = (b[i]-a[i])/sum;
		res+=hm;
		b[i]-=sum*hm;
		trypush((i+1)%n);
		trypush((i-1+n)%n);
	}
	for(int i=0; i<n; ++i){
		if(b[i]!=a[i]){
			cout<<-1<<endl;
			return 0;
		}
	}
	cout <<res<<endl;
	return 0;
}
