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=2e5+5;
typedef pair<int,int>pii;
int n,a[N],b[N],i;
long long su;
priority_queue<pii>q;
inline int lst(int x){
	return x==1?n:x-1;
}
inline int nxt(int x){
	return x==n?1:x+1;
}
int main(){
	scanf("%d",&n);
	for(i=1;i<=n;++i)scanf("%d",a+i);
	for(i=1;i<=n;++i)scanf("%d",b+i),q.push(pii(b[i],i));
	for(;!q.empty();){
		int u=q.top().second;q.pop();
		if(b[u]<a[u]){
			su=-1;break;
		}
		int z=b[lst(u)]+b[nxt(u)];
		if(b[u]%z>a[u]){
			su+=b[u]/z;
			b[u]%=z;q.push(pii(b[u],u));
			if(b[u]>=b[lst(u)] && b[u]>=b[nxt(u)]){
				su=-1;break;
			}
		}else if((b[u]-a[u])%z){
			su=-1;break;
		}else su+=(b[u]-a[u])/z,b[u]=a[u];
	}
	printf("%lld\n",su);
}