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 mp std::make_pair
#define fi first
#define se second
typedef long long LL;
typedef std::pair<int, int> pii;
const int MN = 200005;

int N, A[MN], B[MN];
std::priority_queue<pii> pq;
LL Ans;

inline int nx(const int &x) { return x % N + 1; }
inline int ls(const int &x) { return (x - 2 + N) % N + 1; }

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]);
	for (int i = 1; i <= N; ++i) {
		if (B[i] < A[i]) return puts("-1"), 0;
		if (B[i] > A[i]) pq.push(mp(B[i], i));
	}
	while (!pq.empty()) {
		pii p = pq.top(); pq.pop();
		int i = p.se, v = p.fi;
		int x = B[ls(i)] + B[nx(i)], t = (v - A[i]) / x;
		if (!t) return puts("-1"), 0;
		Ans += t;
		B[i] -= t * x;
		if (B[i] > A[i]) pq.push(mp(B[i], i));
	}
	printf("%lld\n", Ans);
	return 0;
}