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;

int con[300010];
int INF = 1e5;

signed main(){
    //cout << setprecision(16) ;
    ios::sync_with_stdio(false);
	cin.tie(0);

    ll n;
    cin>>n;
    ll a[n],b[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    for(int i=0;i<n;i++){
        cin>>b[i];
    }
    ll ans = 0;
    while(true){
        bool end = true;
        bool ng = true;
        for(int i=0;i<n;i++){
            if(b[i]!=a[i]) end = false;
            if(b[(i+n-1)%n] + b[(i+1)%n] < b[i]){
                ll t = (b[i]-a[i])/(b[(i+n-1)%n] + b[(i+1)%n]);
                ans += t;
                if(t!=0) ng = false;
                b[i] -= t*(b[(i+n-1)%n] + b[(i+1)%n]);
            }
        }
        if(end) break;
        if(ng){
            cout<<-1<<endl;
            return 0;
        }
    }
    cout << ans << endl;

}