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;
#define ii pair<int,int>
#define ll long long

int n;
int arr[200005];
int brr[200005];
priority_queue<ii> pq;

void rage(){
	printf("-1\n");
	exit(0);
}

int main(){
	cin>>n;
	for (int x=0;x<n;x++) cin>>arr[x];
	for (int x=0;x<n;x++) cin>>brr[x];
	
	for (int x=0;x<n;x++){
		if (arr[x]>brr[x]) rage();
		else if (arr[x]<brr[x]) pq.push(ii(brr[x],x));
	}
	
	int pos,temp,diff;
	ll ans=0;
	while (!pq.empty()){
		//printf("%d %d\n",pq.top().first,pq.top().second);
		pos=pq.top().second,pq.pop();
		
		temp=brr[(pos+1)%n]+brr[(pos-1+n)%n];
		diff=brr[pos]-arr[pos];
		
		if (diff<temp) rage();
		ans+=diff/temp;
		brr[pos]=arr[pos]+diff%temp;
		
		if (arr[pos]<brr[pos]){
			pq.push(ii(brr[pos],pos));
		}
	}
	
	printf("%lld\n",ans);
}