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;

#define P(x,y) make_pair(x,y)
#define N 300000
#define lst(x) ((x)==1?n:(x)-1)
#define nxt(x) ((x)==n?1:(x)+1)

int n,a[N],b[N];
long long ans;
priority_queue<pair<int,int> > pq;

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);
		pq.push(P(b[i],i));
	}
	while (!pq.empty()){
		int x=pq.top().second; pq.pop();
		int t=b[lst(x)]+b[nxt(x)];
		if (b[x]%t==a[x]%t){ans+=(b[x]-a[x])/t; b[x]=a[x];}
		else{
			ans+=b[x]/t; b[x]%=t;
			if (b[x]<a[x]){puts("-1"); return 0;}
			if (b[x]>=b[lst(x)]&&b[x]>=b[nxt(x)]){puts("-1"); return 0;}
			pq.push(P(b[x],x));
		}
	}
	printf("%lld\n",ans);
	
	return 0;
}
			