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>

using namespace std;

typedef long long ll;

int a[200005],b[200005];
queue <int> q;

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++) scanf("%d",&b[i]);
  for(int i=1;i<=n;i++) scanf("%d",&a[i]);
  for(int i=1;i<=n;i++) {
  	if (a[i]<b[i]) {
  		puts("-1");
  		return 0;
	  }
    if (a[i]!=b[i]) q.push(i);
  }
  ll ans=0;
  while (!q.empty()) {
  	int x=q.front();q.pop();
  	int l=((x>1)?x-1:n),r=((x<n)?x+1:1);
  	if (a[x]<a[l]+a[r]||a[x]==b[x]) continue;
  	if (b[x]>=a[l]+a[r]&&a[x]%(a[l]+a[r])!=b[x]%(a[l]+a[r])) {
  		puts("-1");
  		return 0;
	  }
	if (b[x]>=a[l]+a[r]) {
		ans+=(a[x]-b[x])/(a[l]+a[r]);
		a[x]=b[x];
	}
	else {
		ans+=a[x]/(a[l]+a[r]);
		a[x]%=(a[l]+a[r]);
		if (a[x]<b[x]) {
			puts("-1");
			return 0;
		}
	}
	q.push(l);q.push(r);
  }
  for(int i=1;i<=n;i++)
    if (a[i]!=b[i]) {
    	puts("-1");
    	return 0;
	}
  printf("%lld\n",ans);
  return 0;
}