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 <iostream>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<ll, int> P;

int main()
{
    int n;
    cin >> n;
    ll a[200005], b[200005];
    for(int i = 0; i < n; i++) cin >> a[i];
    priority_queue<P> que;
    for(int i = 0; i < n; i++){
        cin >> b[i];
        que.push(P(b[i], i));
    }
    ll ans = 0;
    int l = -1;
    while(que.size()){
        P p = que.top();
        que.pop();
        int i = p.second;
        if(i == l){
            cout << -1 << endl;
            return 0;
        }
        l = i;
        ll l = b[(i + n - 1) % n], r = b[(i + 1) % n];
        ll c = (b[i] - a[i]) / (l + r);
        ans += c;
        b[i] -= c * (l + r);
        if(b[i] != a[i]) que.push(P(b[i], i));
    }
    cout << ans << endl;
}
