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;
long long mod =998244353;
int a[200009];
int b[200009];
int main(){
    int n;
    cin >> n;
    set<pair<int,int> > s;
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }
    int k = 0;
    for(int i = 0; i < n; i++){
        cin >> b[i];
        if(b[i] == a[i]) k++;
        else
        s.insert({b[i], i});
    }
    long long t= 0;
    while(s.size()){
        auto it = --s.end();
        int x = it->second;
        s.erase(s.find({b[x],x}));
        int u = b[(x+1)%n], v = b[(x+n-1)%n];
        int m = max(a[x], max(u,v));
        int K = max(1, (b[x] - m + u + v - 1) / (u + v));

        t += K;
        b[x] -= K * (u+v);
        if(b[x] < a[x]){
            cout<<-1<<endl;
            return 0;
        } else if(a[x] != b[x]) s.insert({b[x],x});
      //  t++;

    }
    cout << t << endl;

}
