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 <cstdio>
#include <queue>

using namespace std;

int main()
{
	int n;
	scanf("%d", &n);
	
	int a[200000];
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	
	int b[200000];
	priority_queue<pair<int, int> > pq;
	for (int i = 0; i < n; i++) {
		scanf("%d", &b[i]);
		if (a[i] > b[i]) {
			puts("-1");
			return 0;
		}
		if (a[i] == b[i]) continue;	
		pq.push(make_pair(b[i], i));
	}
	
	long long cnt = 0;
	while (!pq.empty()) {
		int id = pq.top().second;
		pq.pop();
		int s = (id == 0) ? n-1 : id-1;
		int t = (id == n-1) ? 0 : id+1;
		if ((b[id]-a[id])%(b[s]+b[t]) == 0) {
			cnt += (b[id]-a[id])/(b[s]+b[t]);
			b[id] = a[id];
			continue;
		} else if (b[id]-a[id] < b[s]+b[t]) {
			puts("-1");
			return 0;
		} else {
			cnt += b[id]/(b[s]+b[t]);
			b[id] %= b[s]+b[t];
			pq.push(make_pair(b[id], id));
		}
	}
	
	printf("%lld\n", cnt);
	
	return 0;
}