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>
#define pa pair<int,int>
using namespace std;
priority_queue<pa> qwq;
int a[200005],b[200005],N;
int main(){
	scanf("%d",&N);
	long long ans=0;
	for (int i=1;i<=N;i++)
		scanf("%d",&b[i]);
	for (int i=1;i<=N;i++){
		scanf("%d",&a[i]);
		qwq.push(make_pair(a[i],i));
	}
	while (!qwq.empty()){
		pa now=qwq.top();
		qwq.pop();
		if (b[now.second]>now.first) {
			puts("-1");
			return 0;
		}
		if (b[now.second]==now.first) continue;
		int Ned=now.first-b[now.second];
		int tmp=a[now.second-1]+a[now.second+1];
		if (now.second==1) tmp+=a[N];
		if (now.second==N) tmp+=a[1];
		if (tmp>Ned){
			printf("-1\n");
			return 0;
		}
		int tmp1=Ned/tmp;
		ans+=tmp1;
		a[now.second]-=tmp1*tmp;
		now.first=a[now.second];
		qwq.push(now);
	}
	cout<<ans;
	return 0;
}