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;

int main(){
    ll n,ans = 0;
    cin >> n;
    vector<ll> a(n);
    vector<ll> b(n);
    priority_queue<pair<ll,ll>> pq;
    for(int i = 0;i < n;i ++){
        cin >> a[i];
    }
    for(int i = 0;i < n;i ++){
        cin >> b[i];
        pq.push(make_pair(b[i],i));
    }
    while(!pq.empty()){
        ll i = pq.top().second,ln,rn;
        pq.pop();
        ln = (i==0?b[n-1]:b[i-1]);
        rn = (i==n-1?b[0]:b[i+1]);

        if((b[i]-a[i])%(ln+rn)==0){
            ans += (b[i]-a[i])/(ln+rn);
            b[i] = a[i];
        }else if(ln+rn<b[i]){
            ll k = (b[i]%(ln+rn)==0?b[i]/(ln+rn)-1:b[i]/(ln+rn));
            b[i] -= k*(ln+rn);
            ans += k;
            pq.push(make_pair(b[i],i));
        }else{
            cout << -1 << endl;
            return 0;
        }
    }
    cout << ans << endl;
    return 0;
}