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;
const int maxn=2e5+7;
typedef long long ll;
typedef pair<ll,int> pii;
ll a[maxn],b[maxn];
priority_queue<pii> q;
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
	}
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&b[i]);
		if(a[i]!=b[i]) q.push({b[i],i});
	}
	ll ans=0;
	while(!q.empty())
	{
		pii t=q.top();q.pop();
		ll ib=t.first;int id=t.second;
		//if(ib==a[id]) continue;
		int pre = (id + n - 2) % n + 1, suf = id % n + 1;
		int stp=(ib-a[id])/(b[pre]+b[suf]);
		if(!stp)
		{
			puts("-1");return 0;
		}
		ans+=stp;
		ib-=1ll*stp*(b[pre]+b[suf]);
		b[id]=ib;
		if(a[id]!=ib) q.push({ib,id});
	}
	cout<<ans<<endl;
}