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 Rint register int
#define MP make_pair
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int N = 200003;
int n, a[N], b[N];
LL ans;
priority_queue<pii> pq;
int main(){
	scanf("%d", &n);
	for(Rint i = 1;i <= n;i ++) scanf("%d", a + i);
	for(Rint i = 1;i <= n;i ++){
		scanf("%d", b + i);
		if(a[i] != b[i]) pq.push(MP(b[i], i));
	}
	while(!pq.empty()){
		pii now = pq.top(); pq.pop();
		int i = now.se, pre = (now.se + n - 2) % n + 1, suf = now.se % n + 1, step = (b[i] - a[i]) / (b[pre] + b[suf]);
		if(!step){puts("-1"); return 0;}
		ans += step; b[i] -= step * (b[pre] + b[suf]);
		if(a[i] != b[i]) pq.push(MP(b[i], i));
	}
	printf("%lld", ans);
}