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 <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>

#include <queue>

using namespace std;


typedef long long LL;;


typedef pair <int, int> pr;


const int N = 2e5 + 10;


int a[N], b[N];
int n;

priority_queue <pr> q;

int main()
{
	scanf("%d", &n);
	
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
	
	for (int i = 1; i <= n; i++) q.push(make_pair(b[i], i));
	
	LL Ans = 0;
	
	while (!q.empty())
	{
		pr now = q.top(); q.pop();
		int val = now.first, id = now.second;
		if (val == a[id]) continue;
		if (val < a[id]) return 0 & puts("-1");
		int w = val - a[id]; 
		int num = b[id < n ? id + 1 : 1] + b[id > 1 ? id - 1 : n];
		if (num > w) return 0 & puts("-1");
		Ans += w / num;
		b[id] -= (w / num) * num;
		q.push(make_pair(b[id], id));
	}
	
	printf("%lld\n", Ans);
	
	return 0;
}