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 <stdio.h>
#include <queue>
using namespace std;
long long n;
long long a[200000];
long long b[200000];
long long s, t, x, y, z;
long long ans;
priority_queue<long long>q;
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%lld", a + i);
		s += a[i];
	}
	for (int i = 0; i < n; i++) {
		scanf("%lld", b + i);
		t += b[i];
		q.push(b[i] * n + i);
	}
	while (t > s) {
		if (q.empty())break;
		x = q.top();
		x %= n;
		if (b[x] == a[x]) {
			q.pop();
			continue;
		}
		if (b[x] - b[(x + n - 1) % n] - b[(x + 1) % n] < a[x])break;
		q.pop();
		y = b[(x + n - 1) % n] + b[(x + 1) % n];
		z = (b[x] - a[x]) / y;
		b[x] -= z * y;
		t -= z * y;
		ans += z;
		q.push(b[x] * n + x);
	}
	y = 1;
	for (int i = 0; i < n; i++) {
		if (a[i] != b[i])y = 0;
	}
	if (y)printf("%lld\n", ans);
	else printf("-1\n");
}