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 mk make_pair
using namespace std;
typedef pair<int,int> pr;
const int N=200005;

int n,cnt;
int a[N],b[N];
long long ans;
priority_queue<pr>Q;
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]);
		if(b[i]!=a[i])
			Q.push(mk(b[i],i));
	}
	b[0]=b[n]; b[n+1]=b[1];
	while(!Q.empty()) {
		pr now=Q.top(); Q.pop();
		int p=now.second;
		int dir=(b[p]-a[p])/(b[p-1]+b[p+1]);
		if(dir<=0) {
			printf("-1\n");
			return 0;
		}
		b[p]=b[p]-dir*(b[p-1]+b[p+1]);
		if(p==1) b[n+1]=b[p];
		if(p==n) b[0]=b[p];
		ans+=dir;
		if(b[p]!=a[p])
			Q.push(mk(b[p],p));
	}
	printf("%lld\n",ans);
	return 0;
}
