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;

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    priority_queue<pair<int, int>> q;
    for (int i = 0; i < n; i++) {
        cin >> b[i];
        q.push({b[i], i});
    }
    long long ans = 0;
    while (q.size() > 0) {
        int i = q.top().second;
        q.pop();
        if (b[i] < a[i]) {
            ans = -1;
            break;
        }
        if (b[i] > a[i]) {
            int d = b[(n + i - 1) % n] + b[(i + 1) % n];
            int c = (b[i] - a[i]) / d;
            if (c == 0) {
                ans = -1;
                break;
            }
            ans += c;
            b[i] -= c*d;
            q.push({b[i], i});
        }
    }
    cout << ans << endl;
}



