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;

int A[200005];
int B[200005];

typedef pair<int, int> pii;
typedef long long LL;
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];
	}

	priority_queue<pii> pq;
	LL res = 0;
	for (int i = 0; i < N; i++){
		if (A[i] != B[i]){
			pq.push(make_pair(B[i], i));
		}
	}
	while (pq.size()){
		pii p = pq.top(); pq.pop();
		//cout << p.first << ' ' << p.second << endl;
		int idx = p.second;
		int rgt = (idx + 1) % N;
		int lft = (idx + N - 1) % N;
		int dec = (B[idx] - A[idx])  / (B[rgt] + B[lft]);
		if (dec == 0){
			cout << -1 << endl;
			return 0;
		}
		p.first -= dec * (B[rgt] + B[lft]);
		res += dec;
		B[p.second] = p.first;
		if (p.first == A[idx]) continue;
		if (p.first < A[idx]){
			cout << -1 << endl;
			return 0;
		}
		pq.push(p);
	}
	cout << res << endl;
}
