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>
using namespace std;
#define modulo 998244353
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 100000000


int main(){
	
	int N;
	cin>>N;
	
	vector<long long> A(N),B(N);
	for(int i=0;i<N;i++){
		cin>>A[i];
	}
	
	priority_queue<pair<int,int>> P;
	
	for(int i=0;i<N;i++){
		cin>>B[i];
		if(A[i]!=B[i]){
			P.push(make_pair(B[i],i));
		}
	}
	long long ans = 0;
	while(P.size()!=0){
		int a = P.top().first;
		int b = P.top().second;
		P.pop();
		
		long long k = B[(b+1)%N]+B[(b-1+N)%N];
		if(B[b]<A[b]){
			cout<<-1<<endl;
			return 0;
		}
		long long x = B[b]-A[b];
		ans += x/k;
		if(x/k==0){
			cout<<-1<<endl;
			return 0;
		}
		
		B[b] -= (x/k)*k;
		if(A[b]!=B[b]){
			P.push(make_pair(B[b],b));
		}
		
	}
	
	for(int i=0;i<N;i++){
		if(A[i]!=B[i]){
			cout<<-1<<endl;
			return 0;
		}
	}
	
	cout<<ans<<endl;
	
	
	return 0;
}
