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;

typedef pair<int, int> p;

int n, m;
int a[202020], b[202020];

int 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];
	
	priority_queue<p> pq;
	for(int i=0; i<n; i++) pq.push({b[i], i});
	
	long long ans = 0;
	while(pq.size()){
		int i = pq.top().second; pq.pop();
		int st =  b[(i+1)%n] + b[(i+n-1)%n];
		
		if((b[i] - a[i]) % st == 0){
			ans += (b[i] - a[i]) / st;
			b[i] = a[i];
			continue;
		}
		if(st >= b[i]){
			cout << -1; return 0;
		}
		
		ans += b[i] / st;
		b[i] %= st;
		if (b[i] < a[i]) {
			cout << -1; return 0;
		}
		pq.push({b[i], i});
	}
	cout << ans;
}