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<algorithm>
using namespace std;
const int MAXN=200005;
int N,A[MAXN],B[MAXN];
long long Ans;
int Check(){
	for(int i=1;i<=N;i++)
		if(B[i]!=A[i])return 0;
	return 1;
}
int Get(int x){
	if(x==0)return N;
	if(x==N+1)return 1;
	return x;
}
int main(){
	//freopen("txt.in","r",stdin);
	//freopen("hopeless.in","r",stdin);
	//freopen("hopeless.out","w",stdout);
	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]);
	while(!Check()){
		int Ok=0;
		for(int i=1;i<=N;i++)
			if(B[i]-B[Get(i-1)]-B[Get(i+1)]>=A[i]){
				int t1=B[i]-A[i],t2=B[Get(i-1)]+B[Get(i+1)];
				B[i]=t1%t2+A[i];Ans+=t1/t2,Ok=1;
			}
		if(!Ok){
			printf("-1\n");
			return 0;
		}
	}
	printf("%lld\n",Ans);
}
/*
5
5 6 5 2 1
9817 1108 6890 4343 8704

5
1 2 5 6 5
8704 4343 6890 1108 9817

5
1 1 1 1 1
1 1000000001 1 1000000001 1
*/