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 <queue>
#include <utility>
#include <algorithm>
typedef long long ll;
using namespace std;

int A[200010], B[200010];
priority_queue<pair<int, int>> pq;

int main() {
	int N;
	cin >> N;
	for(int i=0; i<N; ++i) cin >> A[i];
	for(int i=0; i<N; ++i){
		cin >> B[i];
		pq.push(make_pair(B[i], i));
	}
	ll ans=0;
	while(pq.size()>0){
		int key=pq.top().second;
		pq.pop();
		if(B[key]<A[key]){
			cout << -1 << endl;
			return 0;
		}
		if(B[key]>A[key]){
			if((B[key]-A[key])%(B[(key+N-1)%N]+B[(key+1)%N])==0){
				ans += (B[key]-A[key])/(B[(key+N-1)%N]+B[(key+1)%N]);
				B[key]=A[key];
			}else{
				ans += B[key]/(B[(key+N-1)%N]+B[(key+1)%N]);
				B[key] %= (B[(key+N-1)%N]+B[(key+1)%N]);
				if(B[key]>=max(B[(key+N-1)%N], B[(key+1)%N])){
					cout << -1 << endl;
					return 0;
				}
				pq.push(make_pair(B[key], key));
			}
		}
	}
	cout << ans << endl;
	return 0;
}