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;
#define ll long long

int main(){
    int N;
    cin >> N;
    vector<ll> A(N), B(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }
    
    priority_queue<pair<ll, int>> pq;
    for (int i = 0; i < N; i++) {
        cin >> B[i];
        if (A[i] < B[i]) {
            pq.push({B[i], i});
        }
        if (A[i] > B[i]) {
            cout << -1 << endl;
            return 0;
        }
    }
    ll ans = 0;
    while (!pq.empty()) {
        int pos = pq.top().second;
        pq.pop();
        int pre = (pos+N-1)%N;
        int nxt = (pos+1)%N;
        ll dist = B[pre]+B[nxt];
        if (dist > B[pos]-A[pos]) {
            cout << -1 << endl;
            return 0;
        }
        ans += (B[pos]-A[pos])/dist;
        B[pos] -= (B[pos]-A[pos])/dist*dist;
        if (B[pos] > A[pos]) pq.push({B[pos], pos});
    }
    cout << ans << endl;
    return 0;
}
