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<iostream>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;

int main(){
    int n;
    cin >> n;
    vector<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];

    priority_queue<pair<int,int>> pq;
    for(int i = 0; i < n; i++){
        if(a[i] > b[i]){
            cout << -1 << endl;
            return 0;
        }else if(a[i] < b[i]){
            pq.push({b[i], i});
        }
    }

    ll ans = 0;
    while(!pq.empty()){
        pair<int,int> p = pq.top(); pq.pop();
        int i = p.second;
        int unit = b[(i-1+n)%n] + b[(i+1)%n];
        if((b[i]-a[i])/unit == 0){
            cout << -1 << endl;
            return 0;
        }
        ans += (b[i]-a[i])/unit;
        b[i] -= (b[i]-a[i])/unit*unit;
        if(b[i] > a[i]) pq.push({b[i], i});
    }

    cout << ans << endl;
    return 0;
}