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 maxn=1e6+34;
typedef pair<int,int> pi;
priority_queue <pi> q;
int a[maxn],b[maxn],n;
ll cnt;

int main(){
    cin >> n;
    for (int i=0;i<n;i++) cin >> a[i];
    for (int i=0;i<n;i++) cin >> b[i];
    for (int i=0;i<n;i++) if (a[i]>b[i]) {puts("-1");return 0;}
    for (int i=0;i<n;i++) if (a[i]!=b[i]) q.push((pi){b[i],i});
    while (!q.empty()){
        pi tmp=q.top(); q.pop();
        int x=tmp.second;
        int temp=(b[x]-a[x])/(b[(x+1)%n]+b[(x+n-1)%n]);
        if (temp==0){puts("-1");return 0;}
        cnt+=temp; b[x]-=temp*(b[(x+1)%n]+b[(x+n-1)%n]);
        if (b[x]>a[x]) q.push((pi){b[x],x});
    }
    cout << cnt << endl;
}
