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+10;
typedef pair<int,int> pii;
#define mp make_pair
#define fi first
#define se second

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,a[N],b[N];
priority_queue<pii> Q;
long long ans=0;

int main() {
	cin>>n;
	for(int i=1;i<=n;i++) a[i]=gi();
	for(int i=1;i<=n;i++) {
		b[i]=gi();
		if(b[i]!=a[i]) Q.push(mp(b[i],i));
		if(b[i]<a[i]) return puts("-1"),0;
	}
	while(!Q.empty()) {
		int i=Q.top().se;Q.pop();
		int x=(i==1?n:i-1),y=(i==n?1:i+1);
		if(b[i]-a[i]<b[x]+b[y]) return puts("-1"),0;
		int d=(b[i]-a[i])/(b[x]+b[y]);
		b[i]-=d*(b[x]+b[y]);ans+=d;
		if(b[i]!=a[i]) Q.push(mp(b[i],i));
	}
	cout<<ans;
	return 0;
}
