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 ll long long
using namespace std;
const int N=2<<17;
int n;ll a[N],b[N],ans;
int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++)scanf("%d",&a[i]);
	for(int i=0;i<n;i++)scanf("%d",&b[i]);
	for(int t=0;t<500;t++)
		for(int i=t&1;i<n;i+=2)
		{
			ll s=b[(i+n-1)%n]+b[(i+1)%n];
			ll t=(b[i]-a[i])/s;
			ans+=t;
			b[i]-=t*s;
		}
	for(int i=0;i<n;i++)
		if(a[i]!=b[i])
		{
			puts("-1");
			return 0;
		}
	printf("%lld\n",ans);
	return 0;
}