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];
priority_queue<pair<ll,ll>> que;
int main() {
	ll n,cnt=0;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	for(int i=0;i<n;i++){
		ll x;
		cin>>x;
		b[i]=x;
		if(a[i]!=b[i]){
		que.push(make_pair(x,i));
		cnt++;
		}
	}
	ll ans=0;
	while(cnt!=0){
		pair<ll,ll> kari=que.top();
		que.pop();
		ll nowval=kari.first;
		ll nowiti=kari.second;
		ll l=nowiti-1,r=nowiti+1;
		if(l<0){
			l=n-1;
		}
		if(r==n){
			r=0;
		}
		ll lval=b[l],rval=b[r];
		ans++;
		nowval-=(lval+rval);
		ll herasu=(nowval-a[nowiti])/(lval+rval);
		ans+=herasu;
		  nowval-=herasu*(lval+rval);
		b[nowiti]=nowval;
		if(nowval<a[nowiti]){
			cout << "-1";
			return 0;
		}
		if(nowval==a[nowiti]){
			cnt--;
		}
		else{
		que.push(make_pair(nowval,nowiti));
		}
	}
	cout << ans;
	// your code goes here
	return 0;
}