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
/*Lucky_Glass*/
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int N=2e5;
#define fir first
#define sec second

int n;
int A[N+3],B[N+3];
bool fail;
priority_queue< pair<int,int> > que;

int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%d",&A[i]);
    for(int i=0;i<n;i++){
        scanf("%d",&B[i]);
        if(B[i]!=A[i]) que.push(make_pair(B[i],i));
        if(B[i]<A[i]) fail=true;
    }
    if(fail) printf("-1\n"),exit(0);
    long long ans=0;
    while(!que.empty()){
        int it=que.top().sec;que.pop();
        int div=B[(it+1)%n]+B[(it-1+n)%n];
        long long delta=(B[it]-A[it])/div;
        if(!delta){fail=true;break;}
        B[it]-=div*delta;
        ans+=delta;
        if(B[it]==A[it]) continue;
        que.push(make_pair(B[it],it));
    }
    if(fail) printf("-1\n");
    else printf("%lld\n",ans);
    return 0;
}
