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>

typedef long long int64;

const int MAX_N = int(2e5) + 10;

void fail() {
	puts("-1");
	exit(0);
}

int n;
int a[MAX_N], b[MAX_N];
std::priority_queue<std::pair<int, int> > q;

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", a + i);
	}
	for (int i = 1; i <= n; ++i) {
		scanf("%d", b + i);
		if (a[i] != b[i])
			q.push(std::make_pair(b[i], i));
	}

	int64 ans = 0;
	while (!q.empty()) {
		auto cur = q.top();
		q.pop();
		int i = cur.second, pre = i == 1 ? n : i - 1, nxt = i == n ? 1 : i + 1;
		int t = (b[i] - a[i]) / (b[pre] + b[nxt]);
		if (t == 0) fail();
		ans += t;
		b[i] -= t * (b[pre] + b[nxt]);
		if (a[i] != b[i])
			q.push(std::make_pair(b[i], i));
	}
	printf("%lld\n", ans);
}