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>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 200000;
int A[MAXN + 5], B[MAXN + 5];
priority_queue<pair<int, int> >que;
int main() {
	int N; scanf("%d", &N);
	for(int i=0;i<N;i++) scanf("%d", &A[i]);
	for(int i=0;i<N;i++) {
		scanf("%d", &B[i]);
		que.push(make_pair(B[i], i));
	}
	long long ans = 0;
	while( !que.empty() ) {
		int x = que.top().first, y = que.top().second; que.pop();
		if( x == A[y] ) continue;
		if( x - (B[(y + N - 1)%N] + B[(y + 1)%N]) < A[y] ) {
			puts("-1");
			return 0;
		}
		int tmp = (x - A[y]) / (B[(y + N - 1)%N] + B[(y + 1)%N]);
		ans += tmp;
		B[y] = B[y] - tmp*(B[(y + N - 1)%N] + B[(y + 1)%N]);
		que.push(make_pair(B[y], y));
	}
	printf("%lld\n", ans);
}