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;
typedef stack<int> S;
const int N=200200;
int n;
int a[N];
int b[N];
S s;
void gg(){
	puts("-1");
	exit(0);
}
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]);
		if(b[i]<a[i]){
			gg();
		}
	}
	for(int i=0;i<n;i++){
		s.push(i);
	}
	LL ans=0;
	while(!s.empty()){
		int x=s.top();
		s.pop();
		int l=(x-1+n)%n;
		int r=(x+1)%n;
		LL db=b[l]+b[r];
		LL da=b[x]-a[x];
		if(b[x]<=db){
			continue;
		}
		int o=da/db;
		if(o==0){
			continue;
		}
		ans+=o;
		b[x]-=o*db;
		assert(b[x]>=a[x]);
		s.push(l);
		s.push(r);
	}
	for(int i=0;i<n;i++){
		if(a[i]!=b[i]){
			gg();
		}
	}
	cout<<ans<<endl;
	return 0;
}