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>

int main() {
    using namespace std;

    unsigned long N;
    cin >> N;

    vector<unsigned long> A(N), B(N);
    copy_n(istream_iterator<unsigned long>(cin), N, A.begin());
    priority_queue<pair<unsigned long, unsigned long>> pq;
    for(unsigned long i = 0; i < N; ++i){
        cin >> B[i];
        if(B[i] != A[i])pq.emplace(B[i], i);
    }

    unsigned long ans{0};
    while(!pq.empty()){
        auto t = pq.top().second;
        pq.pop();
        auto bf = (t + N - 1) % N, nx = (t + 1) % N;
        if(B[t] < B[bf] + B[nx] + A[t])return 0 & puts("-1");
        auto tmp = (B[t] - A[t]) / (B[bf] + B[nx]);
        ans += tmp;
        if((B[t] -= tmp * (B[bf] + B[nx])) > A[t])pq.emplace(B[t], t);
    }

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