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 int long long
const int MOD = 1000000007;

signed main() {
int n;
cin>>n;
int a[n],b[n];
for(int i=0;i<n;i++)cin>>a[i];
for(int i=0;i<n;i++)cin>>b[i];
queue<int> que;
for(int i=0;i<n;i++){
    if(b[i]>a[i]&&(b[i]-a[i])>=b[(n+i-1)%n]+b[(i+1)%n])que.push(i);
}
int ans=0,i;
while(!que.empty()){
    i=que.front();
    que.pop();
    ans+=(b[i]-a[i])/(b[(n+i-1)%n]+b[(i+1)%n]);
    b[i]=a[i]+(b[i]-a[i])%(b[(n+i-1)%n]+b[(i+1)%n]);
    i=(n+i-1)%n;
    if(b[i]>a[i]&&(b[i]-a[i])>=b[(n+i-1)%n]+b[(i+1)%n])que.push(i);
    i=(i+2)%n;
    if(b[i]>a[i]&&(b[i]-a[i])>=b[(n+i-1)%n]+b[(i+1)%n])que.push(i);
}
for(int i=0;i<n;i++){
    if(a[i]!=b[i]){
        cout<<-1;
        return 0;
    }
}
cout<<ans;





return 0;
}
