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 va first
#define vb second
using namespace std;
typedef pair<int,int> pii;
const int MN = 2e5+5;
int N,A[MN],B[MN];
priority_queue<pii> PQ;
int main()
{
    cin >> N;
    for(int i=0; i<N; i++) cin >> A[i];
    for(int i=0; i<N; i++){
        cin >> B[i];
        if(B[i]<A[i]){
            cout << -1;
            return 0;
        }
        if(B[i]>A[i]) PQ.emplace(B[i],i);
    }
    int v,x;
    long long ans=0;
    while(PQ.size()){
        tie(v,x) = PQ.top();
        PQ.pop();
        if(B[x]-(B[(x-1+N)%N]+B[(x+1)%N])<A[x]){
            cout << -1;
            return 0;
        }
        int cnt = (B[x]-A[x])/(B[(x-1+N)%N]+B[(x+1)%N]);
        B[x] -= cnt*(B[(x-1+N)%N]+B[(x+1)%N]);
        ans += cnt;
        if(B[x]>A[x]) PQ.emplace(B[x],x);
    }
    cout << ans;
}
