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>

using namespace std;

const int N = 2e5 + 9;

int n, a[N], b[N];

inline int nxt(int x) {
	return x == n ? 1 : x + 1;
}

inline int lxt(int x) {
	return x == 1 ? n : x - 1;
}

typedef pair<int, int> P;

priority_queue<P> que;

void solve() {
	long long ans = 0;
	for (int i = 1; i <= n; ++i) {
		que.push(make_pair(b[i], i));
	}
	while (!que.empty()) {
		P top = que.top(); que.pop();
		int x = top.second;
		if (a[x] == b[x]) continue;
		int p = b[lxt(x)] + b[nxt(x)];
		if (b[x] - p < a[x]) {
			puts("-1");
			return ;
		} else {
			int d = (b[x] - a[x]) / p;
			ans += d;
			b[x] -= d * p;
			que.push(make_pair(b[x], x));
		}
	}
	printf("%lld\n", ans);
}

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]);
	solve();
	return 0;
}