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 rep(i, m, n) for (int i = m; i < n; ++i)

long long solve() {
    int N; cin >> N;
    vector<int> A(N), B(N);
    rep(i, 0, N) cin >> A[i];
    rep(i, 0, N) cin >> B[i];
    using P = pair<int, int>;
    priority_queue<P> pq;
    long long ans = 0;
    rep(i, 0, N) if(A[i] != B[i]) pq.push(P(B[i], i));
    while(!pq.empty()) {
        P t = pq.top(); pq.pop();
        int i = t.second;
        int v = B[(i + N - 1) % N] + B[(i + 1) % N];
        if((B[i] - A[i]) % v == 0) {
            ans += (B[i] - A[i]) / v;
            B[i] = A[i];
        } else if(B[i] <= v) return -1;
        else {
            ans += (B[i] / v);
            B[i] %= v;
        }
        if(A[i] < B[i]) pq.push(P(B[i], i));
        else if(A[i] > B[i]) return -1;
    }
    return ans;
}

int main() {
    cout << solve() << endl;
    return 0;
}
