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 <iostream>
#include <cstdio>
#include <queue>
#define N 200005
using namespace std;
typedef long long ll;

ll n, ans, a[N], b[N];
priority_queue<pair<ll, ll>> pq;
int main() {
	int i, x, y, z;
	cin >> n;
	for(i=0; i<n; i++) scanf("%lld", &a[i]);
	for(i=0; i<n; i++) {
		scanf("%lld", &b[i]);
		pq.push({b[i], i});
	}
	while(!pq.empty()) {
		i = pq.top().second;
		pq.pop();
		if(b[i] == a[i]) continue;
		if(b[i] < a[i]) return puts("-1")*0;
		y = b[(i-1+n)%n] + b[(i+1)%n];
		z = (b[i] - a[i]) / y;
		if(z==0) return puts("-1")*0;
		ans += z;
		b[i] -= z * y;
		pq.push({b[i], i});
	}
	cout << ans;
	return 0;
}