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;
const int maxn = 200200;
int a[maxn], b[maxn];
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<pair<int,int> > pq;
	for(int i = 0; i < n; i++) if(b[i] > a[i]) pq.push({b[i], i});
	long long ops = 0;
	while(!pq.empty()){
		int tp = pq.top().second; pq.pop();
		int pv = (tp + n - 1) % n;
		int nx = (tp + 1) % n;
		if(b[tp] > b[pv] + b[nx]){
			int upto = (b[tp] - a[tp]) / (b[pv] + b[nx]);
			if(upto == 0){
				cout << -1 << endl;
				return 0;
			}
			ops += upto;
			b[tp] -= upto * (b[pv] + b[nx]);
			if(b[tp] > a[tp]) pq.push({b[tp], tp});
		} else {
			cout << -1 << endl;
			return 0;
		}
	}
	cout << ops << endl;
	return 0;
}
