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 ii pair<int,int>
#define mp make_pair
#define int long long int
using namespace std;

int32_t main(){
	int n,c=0;
	cin>>n;
	vector<int> a(n),b(n);
	for(int i=0;i<n;i++)cin>>a[i];
	for(int i=0;i<n;i++)cin>>b[i];
	
	set<ii> data;
	for(int i=0;i<n;i++){
		data.insert(mp(b[i],i));
	}
	while(!data.empty()){
		int j=data.rbegin()->second;
		data.erase(--data.end());
		int q=b[(j+1)%n]+b[(j-1+n)%n];
		if((b[j]-a[j])%q==0){
			c+=(b[j]-a[j])/q;
			b[j]=a[j];
		}
		else{
			if(b[j]<q){cout<<-1;return 0;}
			c+=b[j]/q;
			b[j]%=q;
			if(b[j]<a[j]){cout<<-1;return 0;}
			data.insert(mp(b[j],j));
		}
	}
	cout<<c;
}
