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;
using Int = int_fast64_t;

Int n, ans;
vector<Int> a, b;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n;
	a.resize(n);
	b.resize(n);
	for(auto &i:a) cin >> i;
	for(auto &i:b) cin >> i;
	using P = pair<Int, Int>;
	priority_queue<P> pq;
	for(size_t i=0; i<a.size(); ++i)
		if(b[i] > a[i])
			pq.push(P(b[i], i));
	while(!pq.empty()){
		P p = pq.top();
		pq.pop();
		Int d = b[(p.second-1+n)%n] + b[(p.second+1)%n];
		Int t = (b[p.second] - 1) / d;
		t = min(t, (b[p.second] - a[p.second]) / d);
		b[p.second] -= t * d;
		ans += t;
//		for(size_t i=0; i<b.size(); ++i)
//			cout << b[i] << " \n"[i+1==b.size()];
		if(t == 0){
			ans = -1;
			break;
		}else if(b[p.second] > a[p.second]){
			pq.push(P(b[p.second], p.second));
		}
	}
	cout << ans << "\n";
}