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> pii;
const int N = 200005;

int n, a[N], b[N];
long long ans;
priority_queue<pii> pq;

int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++) {
		scanf("%d",&a[i]);
	}
	for(int i=0;i<n;i++) {
		scanf("%d",&b[i]);
		pq.push({b[i], i});
	}
	while(!pq.empty()) {
		int A, B;
		tie(A, B) = pq.top();
		pq.pop();
		if(b[B] == a[B]) continue;
		if(b[B] < a[B]) {
			puts("-1");
			return 0;
		}
		int x = b[(B+1)%n] + b[(B+n-1)%n];
		int k = (b[B] - a[B]) / x;
		if(k == 0) {
			puts("-1");
			return 0;
		}
		ans += k;
		b[B] -= x * k;
		pq.push({b[B], B});
	}
	printf("%lld\n", ans);
}
