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
// C.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 2e5 + 200;

typedef pair<int, int> pii;
typedef long long ll;

int n, ai[MAX_N], bi[MAX_N];
priority_queue<pii> pq;
ll ans;

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &ai[i]);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &bi[i]);
		if (bi[i] != ai[i])
			pq.push(make_pair(bi[i], i));
	}
	while (!pq.empty())
	{
		int u = pq.top().second;
		pq.pop();
		int pre = (u + n - 2) % n + 1, suf = u % n + 1, steps = (bi[u] - ai[u]) / (bi[pre] + bi[suf]);
		if (steps == 0)
			puts("-1"), exit(0);
		ans += steps, bi[u] -= steps * (bi[pre] + bi[suf]);
		if (ai[u] != bi[u])
			pq.push(make_pair(bi[u], u));
	}
	printf("%lld\n", ans);
	return 0;
}