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=200005;
int a[N],b[N],l,f[N],n;
void up(int x){
	if (x==1)return;
	if (b[f[x]]>b[f[x/2]]){
		swap(f[x],f[x/2]);
		up(x/2);
	}
} 
void down(int x){
	int i=x;
	if (x*2<=l&&b[f[x]]<b[f[x*2]])i=x*2;
	if (x*2<l&&b[f[i]]<b[f[x*2+1]])i=x*2+1;
	if (i!=x){
		swap(f[x],f[i]);
		down(i);
	}
}
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]),f[i]=i;
	for (int i=1;i<=n;i++)
		if (b[i]<a[i]){
			puts("-1");
			return 0;
		}
	l=n;
	for (int i=n;i;i--)down(i);
	long long ans=0;
	while (l){
		int k=f[1];
		if (b[k]==a[k]){
			f[1]=f[l--];
			down(1);
			continue; 
		}
		int num=b[k==1?n:k-1]+b[k==n?1:k+1];
		int p=(b[k]-a[k])/num;
		if (!p){
			puts("-1");
			return 0;
		}
		b[k]-=p*num;down(1);
		ans+=p;
	} 
	printf("%lld\n",ans);
}