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>
using namespace std;
int f[200050],a[200050],n;
bool inq[200050];
inline int lst(int x){return x-1==0?n:x-1;}
inline int nxt(int x){return x+1>n?1:x+1;}
struct node{
	int p;
	node(){}
	node(int p):p(p){}
};
queue<node> Q;
inline bool check(int i){return a[i]>f[i]&&a[i]-f[i]>=a[lst(i)]+a[nxt(i)];}
int main(){
	int i,t1,t2;
	long long ans=0ll;
	scanf("%d",&n);
	for(i=1;i<=n;++i)scanf("%d",f+i);
	for(i=1;i<=n;++i){
		scanf("%d",a+i);
		if(a[i]<f[i]){
			puts("-1");
			return 0;
		}
	}
	for(i=1;i<=n;++i)if(check(i))Q.push(node(i)),inq[i]=1;
	while(!Q.empty()){
		i=Q.front().p;Q.pop();inq[i]=0;
		t1=a[i]-f[i];
		t2=a[lst(i)]+a[nxt(i)];
		ans+=t1/t2;
		a[i]=f[i]+t1%t2;
		if(!inq[lst(i)]&&check(lst(i)))Q.push(lst(i)),inq[lst(i)]=1;
		if(!inq[nxt(i)]&&check(nxt(i)))Q.push(nxt(i)),inq[nxt(i)]=1;
	}
	for(i=1;i<=n;++i)if(f[i]!=a[i]){
		puts("-1");
		return 0;
	}
	printf("%lld\n",ans);
	return 0;
}