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 MAXNUM 222111
#define rep(i,s,t) for(int i=s;i<t;i++)
#define pii pair<int,int>
typedef long long ll;
int a[MAXNUM],b[MAXNUM];
int main()
{
    int n;scanf("%d",&n);
    rep(i,1,n+1)scanf("%d",&a[i]);
    priority_queue<pii> pq;
    rep(i,1,n+1){
        scanf("%d",&b[i]);
        if(b[i]!=a[i])
            pq.emplace(b[i],i);
    }
    ll res=0;
    while(!pq.empty())
    {
        pii p=pq.top();pq.pop();
        int sum=(b[p.second%n+1]+b[(p.second-2+n)%n+1]),num=max(b[p.second%n+1],b[(p.second-2+n)%n+1]);
        num=max(num,a[p.second]);
        int k=max(1,(b[p.second]-num)/sum);
        b[p.second]-=k*sum,res+=k;
        if(b[p.second]<a[p.second]){res=-1;break;}
        else if(b[p.second]>a[p.second])
            pq.emplace(b[p.second],p.second);
    }
    printf("%lld\n",res);
}