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>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

typedef long long LL;

const int maxn=2e5+5;

struct QST{
	int i,b;
};
bool operator < (const QST &a,const QST &b) {return a.b<b.b;}

int n,a[maxn],b[maxn];
priority_queue<QST> Q;

int main()
{
	scanf("%d",&n);
	fo(i,1,n) scanf("%d",&a[i]);
	fo(i,1,n) scanf("%d",&b[i]);
	
	fo(i,1,n) Q.push((QST){i,b[i]});
	LL ans=0;
	while (!Q.empty())
	{
		QST tp=Q.top(); Q.pop();
		int i=tp.i;
		if (b[i]>a[i] && b[i]>=b[i%n+1]+b[(i==1)?n:i-1])
		{
			int sum=b[i%n+1]+b[(i==1)?n:i-1];
			if (a[i]>=b[i]%sum && (b[i]-a[i])%sum!=0) {ans=-1; break;}
			int d=min(b[i]-a[i],b[i]-b[i]%sum);
			b[i]-=d;
			ans+=d/sum;
			Q.push((QST){i,b[i]});
		}
	}
	fo(i,1,n) if (a[i]!=b[i]) {ans=-1; break;}
	
	printf("%lld\n",ans);
}