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>
#define int long long
using namespace std;
const int NC = 2e5 + 5;
int N, A[NC], B[NC];
int pre(int i) { return (i == 1) ? N : (i - 1); }
int nxt(int i) { return (i == N) ? 1 : (i + 1); }
struct cmp {
	bool operator()(const int &x, const int &y) { return B[x] < B[y]; }
};
priority_queue<int, vector<int>, cmp> H;
signed main() {
	scanf("%lld", &N);
	for (int i = 1; i <= N; ++i) scanf("%lld", &A[i]);
	for (int i = 1; i <= N; ++i) scanf("%lld", &B[i]);
	for (int i = 1; i <= N; ++i) if (A[i] != B[i]) H.push(i);
	int ans = 0;
	while (H.size()) {
		int tp = H.top();
		H.pop();
		int dec = (B[tp] - A[tp]) / (B[pre(tp)] + B[nxt(tp)]);
		if (dec == 0 && B[tp] != A[tp]) {
			printf("-1\n");
			return 0;
		}
		B[tp] -= dec * (B[pre(tp)] + B[nxt(tp)]);
		ans += dec;
		if (B[tp] != A[tp])
			H.push(tp);
	}
	printf("%lld\n", ans);
	return 0;
}