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 n;
int a[200003],b[200003];
queue<int>q;
void check(int i){
	if(i==0) i+=n;
	if(i==n+1) i-=n;
	if(a[i]!=b[i] && b[i]>b[i-1] && b[i]>b[i+1]) q.push(i);
}
int main(){
	ios::sync_with_stdio(false);
	cin >> n;
	for(int i=1; i<=n ;i++) cin >> a[i];
	for(int i=1; i<=n ;i++){
		cin >> b[i];
		if(a[i]>b[i]) return cout << "-1\n",0;
	}
	b[n+1]=b[1];b[0]=b[n];
	for(int i=1; i<=n ;i++){
		check(i);
	}
	ll ans=0;
	while(!q.empty()){
		int i=q.front();q.pop();
		if(b[i]%(b[i-1]+b[i+1])==a[i]%(b[i-1]+b[i+1])){
			ans+=(b[i]-a[i])/(b[i-1]+b[i+1]);
			b[i]=a[i];
		}
		else if(b[i]%(b[i-1]+b[i+1])<a[i]){
			return cout << "-1\n",0;
		}
		else{
			ans+=b[i]/(b[i-1]+b[i+1]);
			b[i]%=(b[i-1]+b[i+1]);
		}
		if(i==n) b[0]=b[n];
		if(i==1) b[n+1]=b[1];
		check(i-1);check(i+1);
	}
	for(int i=1; i<=n ;i++){
		if(a[i]!=b[i]) return cout << "-1\n",0;
	}
	cout << ans << endl;
}