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 rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=200005;
int n;
int a[maxn],b[maxn];
priority_queue<pii>q;
ll ans;

int main(){
	scanf("%d",&n);
	rep(i,n)scanf("%d",&a[i]);
	rep(i,n){
		scanf("%d",&b[i]);
		if(b[i]<a[i]){
			puts("-1");
			return 0;
		}
		if(b[i]>a[i])q.push(mp(b[i],i));
	}
	while(!q.empty()){
		int x=q.top().snd;
		q.pop();
		int tmp=b[x]-a[x],sum=b[(x+1)%n]+b[(x+n-1)%n];
		assert(tmp>0);
		if(tmp<sum){
			puts("-1");
			return 0;
		}
		ans+=tmp/sum;
		b[x]=a[x]+tmp%sum;
		if(b[x]>a[x])q.push(mp(b[x],x));
	}
	printf("%lld\n",ans);
	return 0;
}