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
#define HAVE_STRUCT_TIMESPEC
#include<bits/stdc++.h>
using namespace std;
int a[200007],b[2000007];
priority_queue<pair<int,int> >q;
int n;
int mi(int x){
    return x==1?n:x-1;
}
int pl(int x){
    return x==n?1:x+1;
}
int main(){
    cin>>n;
    for(int i=1;i<=n;++i)
        cin>>a[i];
    for(int i=1;i<=n;++i){
        cin>>b[i];
        q.push({b[i],i});
    }
    long long ans=0;
    while(!q.empty()){
        int i=q.top().second;
        q.pop();
        if(b[i]==a[i])
            continue;
        if(a[i]>b[i]-b[mi(i)]-b[pl(i)])
            return puts("-1"),0;
        int tmp=b[mi(i)]+b[pl(i)];
        int tot=b[i]-a[i];
        ans+=tot/tmp;
        tot%=tmp;
        b[i]=a[i]+tot;
        if(a[i]>b[i])
            return puts("-1"),0;
        if(a[i]!=b[i])
            q.push({b[i],i});
    }
    cout<<ans;
    return 0;
}
