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 N = 2e5+5;
int a[N],b[N],n;
int main(){
    ios::sync_with_stdio(false);
    cin>>n;
    for(int i=0;i<n;i++)cin>>a[i];
    for(int i=0;i<n;i++)cin>>b[i];
    queue<int> q;
    for(int i=0;i<n;i++){
        if(b[i]-a[i]>=b[(i-1+n)%n]+b[(i+1)%n])
            q.push(i);
    }
    ll ans = 0;
    while (!q.empty()){
        int i = q.front();q.pop();
        int k = (b[i]-a[i])/(b[(i+n-1)%n]+b[(i+1)%n]);
        b[i]-=k*(b[(i-1+n)%n]+b[(i+1)%n]);ans+=k;
        i=(i+n-1)%n;
        if(b[i]-a[i]>=b[(i+n-1)%n]+b[(i+1)%n])
            q.push(i);
        i=(i+2)%n;
        if(b[i]-a[i]>=b[(i+n-1)%n]+b[(i+1)%n])
            q.push(i);
    }
    bool f=1;
    for(int i=0;i<n;i++){
        if(a[i]!=b[i]){
            f=0;
            break;
        }
    }
    if(f)cout<<ans<<endl;
    else cout<<-1<<endl;
}