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;
typedef long long ll;

ll n;
ll ans;
ll a[1<<18],b[1<<18];

int pre(int i) {
    return (i + n - 1) % n;
}

int nxt(int i) {
    return (i + 1) % n;
}

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> b[i];
    for (int i = 0; i < n; i++) cin >> a[i];
    priority_queue<pair<ll,ll>> pq;
    for (int i = 0; i < n; i++) {
        pq.push({a[i],i});
    }
    while (!pq.empty()) {
        ll num = pq.top().first;
        ll id = pq.top().second;
        pq.pop();
        ll k = (num - b[id]) / (a[pre(id)] + a[nxt(id)]);
        ans += k;
        a[id] -= k * (a[pre(id)] + a[nxt(id)]);
        if (a[id] == b[id]) continue;
        if (k == 0 && a[id] != b[id]) {
            cout << -1 << endl;
            return 0;
        }
        pq.push({a[id], id});
    }

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