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 all(x) (x).begin(), (x).end()
#define mp make_pair
#define pb push_back
#define ff first
#define ss second
#define ll long long
typedef pair<ll,ll> pll;
typedef vector <pll> vi;
#define out(x) return cout<<x,0;
#define precision(x,d) cout<<fixed<<setprecision(d)<<x
#define coutif(a,b,c) cout<<(a?b:c)

int main(){
    ll n,m,j,k=0,ans=0,t=0,x,y,i;
    cin>>n;
    ll a[n],b[n];
    for(i=0;i<n;i++) cin>>a[i];
    for(i=0;i<n;i++) cin>>b[i];
    for(j=0;j<100;j++){
        for(i=0;i<n;i++){
            x=(i+n-1)%n,y=(i+1)%n;
            k=b[x]+b[y];
            if(k<b[i]){
                t=(b[i]-a[i])/k;
                b[i]-=k*t; 
                ans+=t;
            }
        }
    }
    for(i=0;i<n;i++) if(a[i]!=b[i]) out(-1);
    cout<<ans;
}