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<iostream>
#include<queue>
#define int long long
using namespace std;
priority_queue<pair<int,int> > lis;
const int N=2e5+2;
int ar[N],ar1[N];
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,i,j,k,l,z,val,mod,num=0;
	cin>>n;
	for(i=1;i<=n;i++){
		cin>>ar[i];
	}	
	for(i=1;i<=n;i++){
		cin>>ar1[i];
		if(ar1[i]<ar[i]){
			cout<<-1;
			return 0;
		}
		if(ar1[i]!=ar[i]){
			lis.push({ar1[i],i});
		}
	}
	while(lis.size()){
		j=lis.top().first;
		k=lis.top().second;
		lis.pop();
		if(k==1){
			l=n;
		}
		else{
			l=k-1;
		}
		z=k%n+1;
		mod=ar1[z]+ar1[l];
		val=max(ar[k],j%mod);
		i=(j-val)/mod;
		if(i==0){
			cout<<-1;
			return 0;
		}
		j-=i*mod;
		num+=i;
		ar1[k]=j;
		if(j!=ar[k]){
			lis.push({j,k});
		}
	}
	cout<<num;
}