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
/*input
3
1 1 1
13 5 7
*/
#include "bits/stdc++.h"
using namespace std;
long long a[200005],b[200005],n;

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin >> n;
	priority_queue<pair<long long,int>> pq;
	for(int i=0;i<n;++i) cin >> a[i];
	for(int i=0;i<n;++i) {
		cin >> b[i];
		if(b[i]>a[i]) pq.push({b[i],i});
		else if(b[i]<a[i]) return cout<<-1,0;
	}

	long long ans = 0;
	int iter = 0;
	while(pq.size()){
		auto i = pq.top().second; pq.pop();
		long long left = b[i?i-1:n-1], right = b[i<n-1?i+1:0];
		long long mn = min(left,right);
		long long sum = left+right;

		long long moves = (b[i]-max({left,right,a[i]})+sum-1)/sum;
		ans += moves;
		b[i] -= sum*moves;
		if(b[i]>a[i]) pq.push({b[i],i});
		else if(b[i]<a[i])
			return cout<<-1,0;
		// if(iter++==25) break;
		// for(int i=0;i<n;cout<<b[i++]<<' '); cout << '\n';
	}
	cout << ans;
}