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;
#define rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;
int main(){
    ll n,A[200001],B[200001];
    cin>>n;
    rep(i,n) cin>>A[i];
    rep(i,n) cin>>B[i];
    ll C[200001]={0};
    rep(i,n) C[i]=B[(i-1+n)%n]+B[(i+1)%n];
    queue<int> que;
    rep(i,n) if(C[i]<=B[i]) que.push(i);
    ll ans=0;
    while(!que.empty()){
        int a=que.front();
        que.pop();
        if(C[a]>B[a]||A[a]==B[a]) continue;
        ll b=(B[a]-A[a])/C[a];
        B[a]-=b*C[a];
        C[(a-1+n)%n]-=b*C[a];
        C[(a+1)%n]-=b*C[a];
        ans+=b;
        if(C[(a-1+n)%n]<=B[(a-1+n)%n]) que.push((a-1+n)%n);
        if(C[(a+1)%n]<=B[(a+1)%n]) que.push((a+1)%n);
    }
    int x=1;
    rep(i,n) if(A[i]!=B[i]) x=0;
    if(x) cout<<ans;
    else cout<<-1;
}