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<algorithm>
#include<set>
#include<queue>
using namespace std;
#define pii pair<int,int>
int n, A[201000], B[201000];
priority_queue<pii>PQ;
int main() {
	int i;
	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d", &A[i]);
	}
	for (i = 0; i < n; i++) {
		scanf("%d", &B[i]);
		PQ.push({ B[i],i });
	}
	long long res = 0;
	while (!PQ.empty()) {
		pii it = PQ.top();
		PQ.pop();
		int pv = it.second;
		if (B[pv] <= A[pv]) {
			continue;
		}
		long long aa = B[(pv + n - 1) % n];
		long long bb = A[pv];
		long long cc = B[(pv + 1) % n];
		if (B[pv] < aa + bb + cc) {
			puts("-1");
			return 0;
		}
		long long z = (B[pv] - bb) / (aa + cc);
		res += z;
		B[pv] -= z*(aa + cc);
		if (A[pv] != B[pv]) {
			PQ.push({ B[pv],pv });
		}
	}
	for (i = 0; i < n; i++) {
		if (A[i] != B[i]) {
			puts("-1");
			return 0;
		}
	}
	printf("%lld\n", res);
}