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 <queue>
#include <iostream>
using namespace std;
int main() {
    int n; cin >> n;
    int a[n], b[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    priority_queue<pair<int, int> > pq;
    for (int i = 0; i < n; i++) {
        cin >> b[i];
        if (a[i] > b[i]) { cout << "-1\n"; return 0; }
        if (a[i] < b[i]) pq.push({b[i], i});
    }
    long long ans = 0;
    while (!pq.empty()) {
        int i = pq.top().second;
        pq.pop();
        int l = (i+n-1)%n, r = (i+1)%n;
        if (b[l] + b[r] > b[i] - a[i]) { cout << "-1\n"; return 0; }
        int k = (b[i] - a[i]) / (b[l] + b[r]);
        ans += k;
        b[i] -= k * (b[l] + b[r]);
        if (a[i] < b[i]) pq.push({b[i], i});
    }
    cout << ans << endl;
}
