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 <cstdio>
#include <queue>
#define N 200005
using namespace std;

int n , s[N] , t[N];
typedef pair<int , int> pa;

priority_queue<pa> q;

int prev(int x){return x == 1 ? n : x - 1;}
int next(int x){return x == n ? 1 : x + 1;}
long long ans;
int main()
{
	scanf("%d" , &n);
	for(int i = 1 ; i <= n ; i ++) 
		scanf("%d" , t + i);
	for(int i = 1 ; i <= n ; i ++)
		scanf("%d" , s + i) , q.push( make_pair(s[i] , i) );
	while(!q.empty())
	{
		int val = q.top().first , id = q.top().second;
		q.pop();
		if(t[id] == val)  continue;
		int around = s[prev(id)] + s[next(id)];
		int cnt = (val - t[id]) / around;
		if(cnt <= 0) return puts("-1") & 0;
		ans += cnt; s[id] = val - cnt * around;
		q.push( make_pair(s[id] , id) );
	}
	return printf("%lld\n" , ans) & 0;
}