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>
#define rep(i, n) for (ll i = 0; i < (n); i++)
using ll = long long;
using namespace std;

int main(){
    ll n; cin >> n;
    vector<ll> a(n), b(n);
    rep(i,n) cin >> a[i];
    rep(i,n) cin >> b[i];

    priority_queue<pair<ll,ll>,vector<pair<ll,ll>>,less<pair<ll,ll>>> que;
    ll count = 0;
    rep(i,n){ if(b[i] > a[i]) que.push(make_pair(b[i],i)); else if(b[i] == a[i]) count++; }
  
    ll ans = 0;
    while(!que.empty()){
      	pair<ll,ll> p = que.top(); que.pop();
        ll num, id; tie(num,id) = p;
        ll div = b[(id+1)%n] + b[(id+n-1)%n];
        ll k = (num - a[id]) / div;
        ans += k; if(k == 0){ cout << -1 << endl; return 0; }
        b[id] -= k * div; 
        if(b[id] > a[id]) que.push(make_pair(b[id],id));
        else if( b[id] == a[id]) count++;
    }
    cout << (count == n ? ans : -1) << endl;
    return 0;
}