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
// In the name of God

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2*(int)1e5+5;
int a[N], b[N], n;
priority_queue<pair<int,int>> pq;
signed main() {
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n;
	for(int i=0;i<n;i++) cin >> a[i];
	for(int i=0;i<n;i++) cin >> b[i];
	for(int i=0;i<n;i++) pq.push(make_pair(b[i], i));
	int ans = 0;
	int last = -1;
	while(!pq.empty()) {
		auto tp = pq.top();
		pq.pop();
		if(tp.second==last) {
			cout << -1;
			exit(0);
		}
		int l = tp.second==0?b[n-1]:b[tp.second-1];
		int r = tp.second==n-1?b[0]:b[tp.second+1];
		ans += (tp.first)/(l+r);
		b[tp.second]%=(l+r);
		while(b[tp.second]<a[tp.second]) {
			b[tp.second]+=l+r;
			ans--;
		}
		if(b[tp.second]!=a[tp.second]) pq.push(make_pair(b[tp.second], tp.second));
		last=tp.second;
	}
	cout << ans;
	return 0;
}
