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;
typedef long long ll;
const int max_n=200005;
int n,a[max_n],b[max_n];
struct Node{
    int x;
    bool operator < (Node p)const {
        return b[x]<b[p.x];
    }
};
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);
    for(int i=0;i<n;i++){
        if(b[i]<a[i]){
            printf("-1\n");
            return 0;
        }
    }
    priority_queue<Node> que;
    for(int i=0;i<n;i++)if(b[i]>a[i])que.push({i});
    ll ans=0;
    while(!que.empty()){
        Node p=que.top();que.pop();
        int x=p.x;
        int k=b[(x+1)%n]+b[(x+n-1)%n];
        int delt=b[x]-a[x];
        if(delt<k){
            printf("-1\n");
            return 0;
        }
        ans+=delt/k;
        b[x]=a[x]+delt%k;
        if(b[x]!=a[x])que.push({x});
    }
    printf("%lld\n",ans);
    return 0;
}