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;
#define INF 1LL<<62
#define inf 1000000007

ll a[200010],b[200010];
int main() {
	ll n;
	cin>>n;
	for(ll i=0;i<n;i++){
		cin>>a[i];
	}
	priority_queue<pair<ll,ll>>ch;
	for(ll i=0;i<n;i++){
		cin>>b[i];
		ch.push(make_pair(b[i],i));
	}
	ll ans=0;
	while(!ch.empty()){
		ll val=ch.top().first;
		ll i=ch.top().second;
		ch.pop();
		ll l=i-1;
		if(l<0){
			l=n-1;
		}
		ll r=i+1;
		if(r==n){
			r=0;
		}
		if(val<a[i]){
			cout << -1;
			return 0;
		}
		if(val==a[i]){
			continue;
		}
		ll m=((val)-a[i])/(b[l]+b[r]);
		if(m==0){
			cout << -1;
			return 0;
		}
		ans+=m;
		val=val-m*(b[l]+b[r]);
		ch.push(make_pair(val,i));
		b[i]=val;
	}
	cout <<ans;
	// your code goes here
	return 0;
}